Evaluate the following limits:  

\displaystyle \textbf{Question 1: } \lim \limits_{x \to 0} \frac{\sqrt{1+x+x^2}-1}{x}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=0,\text{ the expression }\frac{\sqrt{1+x+x^2}-1}{x}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Rationalizing the numerator, we get}
\displaystyle \lim \limits_{x \to 0}\left(\frac{\sqrt{1+x+x^2}-1}{x}\right)
\displaystyle =\lim \limits_{x \to 0}\left(\frac{(\sqrt{1+x+x^2}-1)(\sqrt{1+x+x^2}+1)}{x(\sqrt{1+x+x^2}+1)}\right)
\displaystyle =\lim \limits_{x \to 0}\left(\frac{1+x+x^2-1}{x(\sqrt{1+x+x^2}+1)}\right)
\displaystyle =\lim \limits_{x \to 0}\left(\frac{x(1+x)}{x(\sqrt{1+x+x^2}+1)}\right)
\displaystyle =\lim \limits_{x \to 0}\left(\frac{1+x}{\sqrt{1+x+x^2}+1}\right)
\displaystyle =\frac{1+0}{\sqrt{1+0+0}+1}
\displaystyle =\frac{1}{2}
\displaystyle \\

\displaystyle \textbf{Question 2: } \lim \limits_{x \to 0}\frac{2x}{\sqrt{a+x}-\sqrt{a-x}},\quad a>0.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=0,\text{ the expression }\frac{2x}{\sqrt{a+x}-\sqrt{a-x}}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Rationalizing the denominator, we get}
\displaystyle \lim \limits_{x \to 0}\frac{2x}{\sqrt{a+x}-\sqrt{a-x}}
\displaystyle =\lim \limits_{x \to 0}\frac{2x}{\sqrt{a+x}-\sqrt{a-x}}\times\frac{\sqrt{a+x}+\sqrt{a-x}}{\sqrt{a+x}+\sqrt{a-x}}
\displaystyle =\lim \limits_{x \to 0}\frac{2x(\sqrt{a+x}+\sqrt{a-x})}{(a+x)-(a-x)}
\displaystyle =\lim \limits_{x \to 0}\frac{2x(\sqrt{a+x}+\sqrt{a-x})}{2x}
\displaystyle =\lim \limits_{x \to 0}\left(\sqrt{a+x}+\sqrt{a-x}\right)
\displaystyle =\sqrt{a}+\sqrt{a}
\displaystyle =2\sqrt{a}
\displaystyle \\

\displaystyle \textbf{Question 3: } \lim \limits_{x \to 0}\frac{\sqrt{a^2+x^2}-a}{x^2},\quad a>0.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=0,\text{ the expression }\frac{\sqrt{a^2+x^2}-a}{x^2}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Rationalizing the numerator, we get}
\displaystyle \lim \limits_{x \to 0}\frac{\sqrt{a^2+x^2}-a}{x^2}
\displaystyle =\lim \limits_{x \to 0}\frac{(\sqrt{a^2+x^2}-a)(\sqrt{a^2+x^2}+a)}{x^2(\sqrt{a^2+x^2}+a)}
\displaystyle =\lim \limits_{x \to 0}\frac{a^2+x^2-a^2}{x^2(\sqrt{a^2+x^2}+a)}
\displaystyle =\lim \limits_{x \to 0}\frac{x^2}{x^2(\sqrt{a^2+x^2}+a)}
\displaystyle =\lim \limits_{x \to 0}\frac{1}{\sqrt{a^2+x^2}+a}
\displaystyle =\frac{1}{\sqrt{a^2}+a}
\displaystyle =\frac{1}{a+a}
\displaystyle =\frac{1}{2a}
\displaystyle \\

\displaystyle \textbf{Question 4: } \lim \limits_{x \to 0}\frac{\sqrt{1+x}-\sqrt{1-x}}{2x}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=0,\text{ the expression }\frac{\sqrt{1+x}-\sqrt{1-x}}{2x}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Rationalizing the numerator, we get}
\displaystyle \lim \limits_{x \to 0}\frac{\sqrt{1+x}-\sqrt{1-x}}{2x}
\displaystyle =\lim \limits_{x \to 0}\frac{(\sqrt{1+x}-\sqrt{1-x})(\sqrt{1+x}+\sqrt{1-x})}{2x(\sqrt{1+x}+\sqrt{1-x})}
\displaystyle =\lim \limits_{x \to 0}\frac{(1+x)-(1-x)}{2x(\sqrt{1+x}+\sqrt{1-x})}
\displaystyle =\lim \limits_{x \to 0}\frac{2x}{2x(\sqrt{1+x}+\sqrt{1-x})}
\displaystyle =\lim \limits_{x \to 0}\frac{1}{\sqrt{1+x}+\sqrt{1-x}}
\displaystyle =\frac{1}{\sqrt{1+0}+\sqrt{1-0}}
\displaystyle =\frac{1}{2}
\displaystyle \\

\displaystyle \textbf{Question 5: } \lim \limits_{x \to 2}\frac{\sqrt{3-x}-1}{2-x}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=2,\text{ the expression }\frac{\sqrt{3-x}-1}{2-x}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Rationalizing the numerator, we get}
\displaystyle \lim \limits_{x \to 2}\frac{\sqrt{3-x}-1}{2-x}
\displaystyle =\lim \limits_{x \to 2}\frac{(\sqrt{3-x}-1)(\sqrt{3-x}+1)}{(2-x)(\sqrt{3-x}+1)}
\displaystyle =\lim \limits_{x \to 2}\frac{(3-x)-1}{(2-x)(\sqrt{3-x}+1)}
\displaystyle =\lim \limits_{x \to 2}\frac{2-x}{(2-x)(\sqrt{3-x}+1)}
\displaystyle =\lim \limits_{x \to 2}\frac{1}{\sqrt{3-x}+1}
\displaystyle =\frac{1}{\sqrt{3-2}+1}
\displaystyle =\frac{1}{2}
\displaystyle \\

\displaystyle \textbf{Question 6: } \lim \limits_{x \to 3}\frac{x-3}{\sqrt{x-2}-\sqrt{4-x}}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=3,\text{ the expression }\frac{x-3}{\sqrt{x-2}-\sqrt{4-x}}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Rationalizing the denominator, we get}
\displaystyle \lim \limits_{x \to 3}\frac{x-3}{\sqrt{x-2}-\sqrt{4-x}}
\displaystyle =\lim \limits_{x \to 3}\frac{(x-3)(\sqrt{x-2}+\sqrt{4-x})}{(\sqrt{x-2}-\sqrt{4-x})(\sqrt{x-2}+\sqrt{4-x})}
\displaystyle =\lim \limits_{x \to 3}\frac{(x-3)(\sqrt{x-2}+\sqrt{4-x})}{(x-2)-(4-x)}
\displaystyle =\lim \limits_{x \to 3}\frac{(x-3)(\sqrt{x-2}+\sqrt{4-x})}{2x-6}
\displaystyle =\lim \limits_{x \to 3}\frac{(x-3)(\sqrt{x-2}+\sqrt{4-x})}{2(x-3)}
\displaystyle =\lim \limits_{x \to 3}\frac{\sqrt{x-2}+\sqrt{4-x}}{2}
\displaystyle =\frac{\sqrt{3-2}+\sqrt{4-3}}{2}
\displaystyle =\frac{\sqrt{1}+\sqrt{1}}{2}
\displaystyle =1
\displaystyle \\

\displaystyle \textbf{Question 7: } \lim \limits_{x \to 0}\frac{x}{\sqrt{1+x}-\sqrt{1-x}}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=0,\text{ the expression }\frac{x}{\sqrt{1+x}-\sqrt{1-x}}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Rationalizing the denominator, we get}
\displaystyle \lim \limits_{x \to 0}\frac{x}{\sqrt{1+x}-\sqrt{1-x}}
\displaystyle =\lim \limits_{x \to 0}\frac{x(\sqrt{1+x}+\sqrt{1-x})}{(\sqrt{1+x}-\sqrt{1-x})(\sqrt{1+x}+\sqrt{1-x})}
\displaystyle =\lim \limits_{x \to 0}\frac{x(\sqrt{1+x}+\sqrt{1-x})}{(1+x)-(1-x)}
\displaystyle =\lim \limits_{x \to 0}\frac{x(\sqrt{1+x}+\sqrt{1-x})}{2x}
\displaystyle =\lim \limits_{x \to 0}\frac{\sqrt{1+x}+\sqrt{1-x}}{2}
\displaystyle =\frac{\sqrt{1+0}+\sqrt{1-0}}{2}
\displaystyle =\frac{\sqrt{1}+\sqrt{1}}{2}
\displaystyle =1
\displaystyle \\

\displaystyle \textbf{Question 8: } \lim \limits_{x \to 1}\frac{\sqrt{5x-4}-\sqrt{x}}{x-1}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=1,\text{ the expression }\frac{\sqrt{5x-4}-\sqrt{x}}{x-1}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Rationalizing the numerator, we get}
\displaystyle \lim \limits_{x \to 1}\frac{\sqrt{5x-4}-\sqrt{x}}{x-1}
\displaystyle =\lim \limits_{x \to 1}\frac{(\sqrt{5x-4}-\sqrt{x})(\sqrt{5x-4}+\sqrt{x})}{(x-1)(\sqrt{5x-4}+\sqrt{x})}
\displaystyle =\lim \limits_{x \to 1}\frac{5x-4-x}{(x-1)(\sqrt{5x-4}+\sqrt{x})}
\displaystyle =\lim \limits_{x \to 1}\frac{4(x-1)}{(x-1)(\sqrt{5x-4}+\sqrt{x})}
\displaystyle =\lim \limits_{x \to 1}\frac{4}{\sqrt{5x-4}+\sqrt{x}}
\displaystyle =\frac{4}{\sqrt{5-4}+\sqrt{1}}
\displaystyle =\frac{4}{2}
\displaystyle =2
\displaystyle \\

\displaystyle \textbf{Question 9: } \lim \limits_{x \to 1}\frac{x-1}{\sqrt{x^2+3}-2}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=1,\text{ the expression }\frac{x-1}{\sqrt{x^2+3}-2}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Rationalizing the denominator, we get}
\displaystyle \lim \limits_{x \to 1}\frac{x-1}{\sqrt{x^2+3}-2}
\displaystyle =\lim \limits_{x \to 1}\frac{(x-1)(\sqrt{x^2+3}+2)}{(\sqrt{x^2+3}-2)(\sqrt{x^2+3}+2)}
\displaystyle =\lim \limits_{x \to 1}\frac{(x-1)(\sqrt{x^2+3}+2)}{x^2+3-4}
\displaystyle =\lim \limits_{x \to 1}\frac{(x-1)(\sqrt{x^2+3}+2)}{x^2-1}
\displaystyle =\lim \limits_{x \to 1}\frac{(x-1)(\sqrt{x^2+3}+2)}{(x-1)(x+1)}
\displaystyle =\lim \limits_{x \to 1}\frac{\sqrt{x^2+3}+2}{x+1}
\displaystyle =\frac{\sqrt{1^2+3}+2}{1+1}
\displaystyle =\frac{2+2}{2}
\displaystyle =\frac{4}{2}
\displaystyle =2
\displaystyle \\

\displaystyle \textbf{Question 10: } \lim \limits_{x \to 3}\frac{\sqrt{x+3}-\sqrt{6}}{x^2-9}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=3,\text{ the expression }\frac{\sqrt{x+3}-\sqrt{6}}{x^2-9}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Rationalizing the numerator, we get}
\displaystyle \lim \limits_{x \to 3}\frac{\sqrt{x+3}-\sqrt{6}}{x^2-9}
\displaystyle =\lim \limits_{x \to 3}\frac{(\sqrt{x+3}-\sqrt{6})(\sqrt{x+3}+\sqrt{6})}{(x^2-9)(\sqrt{x+3}+\sqrt{6})}
\displaystyle =\lim \limits_{x \to 3}\frac{x+3-6}{(x-3)(x+3)(\sqrt{x+3}+\sqrt{6})}
\displaystyle =\lim \limits_{x \to 3}\frac{x-3}{(x-3)(x+3)(\sqrt{x+3}+\sqrt{6})}
\displaystyle =\lim \limits_{x \to 3}\frac{1}{(x+3)(\sqrt{x+3}+\sqrt{6})}
\displaystyle =\frac{1}{(3+3)(\sqrt{3+3}+\sqrt{6})}
\displaystyle =\frac{1}{6(\sqrt{6}+\sqrt{6})}
\displaystyle =\frac{1}{12\sqrt{6}}
\displaystyle \\

\displaystyle \textbf{Question 11: } \lim \limits_{x \to 1}\frac{\sqrt{5x-4}-\sqrt{x}}{x^2-1}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=1,\text{ the expression }\frac{\sqrt{5x-4}-\sqrt{x}}{x^2-1}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Rationalizing the numerator, we get}
\displaystyle \lim \limits_{x \to 1}\frac{\sqrt{5x-4}-\sqrt{x}}{x^2-1}
\displaystyle =\lim \limits_{x \to 1}\frac{(\sqrt{5x-4}-\sqrt{x})(\sqrt{5x-4}+\sqrt{x})}{(x^2-1)(\sqrt{5x-4}+\sqrt{x})}
\displaystyle =\lim \limits_{x \to 1}\frac{5x-4-x}{(x-1)(x+1)(\sqrt{5x-4}+\sqrt{x})}
\displaystyle =\lim \limits_{x \to 1}\frac{4(x-1)}{(x-1)(x+1)(\sqrt{5x-4}+\sqrt{x})}
\displaystyle =\lim \limits_{x \to 1}\frac{4}{(x+1)(\sqrt{5x-4}+\sqrt{x})}
\displaystyle =\frac{4}{(1+1)(\sqrt{5-4}+\sqrt{1})}
\displaystyle =\frac{4}{2(1+1)}
\displaystyle =\frac{4}{4}
\displaystyle =1
\displaystyle \\

\displaystyle \textbf{Question 12: } \lim \limits_{x \to 0}\frac{\sqrt{1+x}-1}{x}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=0,\text{ the expression }\frac{\sqrt{1+x}-1}{x}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Rationalizing the numerator, we get}
\displaystyle \lim \limits_{x \to 0}\frac{\sqrt{1+x}-1}{x}
\displaystyle =\lim \limits_{x \to 0}\frac{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}{x(\sqrt{1+x}+1)}
\displaystyle =\lim \limits_{x \to 0}\frac{1+x-1}{x(\sqrt{1+x}+1)}
\displaystyle =\lim \limits_{x \to 0}\frac{x}{x(\sqrt{1+x}+1)}
\displaystyle =\lim \limits_{x \to 0}\frac{1}{\sqrt{1+x}+1}
\displaystyle =\frac{1}{\sqrt{1+0}+1}
\displaystyle =\frac{1}{2}
\displaystyle \\

\displaystyle \textbf{Question 13: } \lim \limits_{x \to 2}\frac{\sqrt{x^2+1}-\sqrt{5}}{x-2}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=2,\text{ the expression }\frac{\sqrt{x^2+1}-\sqrt{5}}{x-2}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Rationalizing the numerator, we get}
\displaystyle \lim \limits_{x \to 2}\frac{\sqrt{x^2+1}-\sqrt{5}}{x-2}
\displaystyle =\lim \limits_{x \to 2}\frac{(\sqrt{x^2+1}-\sqrt{5})(\sqrt{x^2+1}+\sqrt{5})}{(x-2)(\sqrt{x^2+1}+\sqrt{5})}
\displaystyle =\lim \limits_{x \to 2}\frac{x^2+1-5}{(x-2)(\sqrt{x^2+1}+\sqrt{5})}
\displaystyle =\lim \limits_{x \to 2}\frac{x^2-4}{(x-2)(\sqrt{x^2+1}+\sqrt{5})}
\displaystyle =\lim \limits_{x \to 2}\frac{(x-2)(x+2)}{(x-2)(\sqrt{x^2+1}+\sqrt{5})}
\displaystyle =\lim \limits_{x \to 2}\frac{x+2}{\sqrt{x^2+1}+\sqrt{5}}
\displaystyle =\frac{2+2}{\sqrt{2^2+1}+\sqrt{5}}
\displaystyle =\frac{4}{2\sqrt{5}}
\displaystyle =\frac{2}{\sqrt{5}}
\displaystyle \\

\displaystyle \textbf{Question 14: } \lim \limits_{x \to 2}\frac{x-2}{\sqrt{x}-\sqrt{2}}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=2,\text{ the expression }\frac{x-2}{\sqrt{x}-\sqrt{2}}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Rationalizing the denominator, we get}
\displaystyle \lim \limits_{x \to 2}\frac{x-2}{\sqrt{x}-\sqrt{2}}
\displaystyle =\lim \limits_{x \to 2}\frac{(\sqrt{x})^2-(\sqrt{2})^2}{\sqrt{x}-\sqrt{2}}
\displaystyle =\lim \limits_{x \to 2}\frac{(\sqrt{x}-\sqrt{2})(\sqrt{x}+\sqrt{2})}{\sqrt{x}-\sqrt{2}}
\displaystyle =\lim \limits_{x \to 2}\left(\sqrt{x}+\sqrt{2}\right)
\displaystyle =\sqrt{2}+\sqrt{2}
\displaystyle =2\sqrt{2}
\displaystyle \\

\displaystyle \textbf{Question 15: } \lim \limits_{x \to 7}\frac{4-\sqrt{9+x}}{1-\sqrt{8-x}}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=7,\text{ the expression }\frac{4-\sqrt{9+x}}{1-\sqrt{8-x}}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Rationalizing the numerator and denominator, we get}
\displaystyle \lim \limits_{x \to 7}\frac{4-\sqrt{9+x}}{1-\sqrt{8-x}}
\displaystyle =\lim \limits_{x \to 7}\frac{4-\sqrt{9+x}}{1-\sqrt{8-x}}\times\frac{4+\sqrt{9+x}}{4+\sqrt{9+x}}\times\frac{1+\sqrt{8-x}}{1+\sqrt{8-x}}
\displaystyle =\lim \limits_{x \to 7}\frac{\left[16-(9+x)\right]\left(1+\sqrt{8-x}\right)}{\left(4+\sqrt{9+x}\right)\left[1-(8-x)\right]}
\displaystyle =\lim \limits_{x \to 7}\frac{(7-x)\left(1+\sqrt{8-x}\right)}{(4+\sqrt{9+x})(x-7)}
\displaystyle =\lim \limits_{x \to 7}\frac{-(x-7)\left(1+\sqrt{8-x}\right)}{(4+\sqrt{9+x})(x-7)}
\displaystyle =\lim \limits_{x \to 7}-\frac{1+\sqrt{8-x}}{4+\sqrt{9+x}}
\displaystyle =-\frac{1+\sqrt{8-7}}{4+\sqrt{9+7}}
\displaystyle =-\frac{1+1}{4+4}
\displaystyle =-\frac{2}{8}
\displaystyle =-\frac{1}{4}
\displaystyle \\

\displaystyle \textbf{Question 16: } \lim \limits_{x \to 0}\frac{\sqrt{a+x}-\sqrt{a}}{x\sqrt{a^2+ax}},\quad a>0.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=0,\text{ the expression }\frac{\sqrt{a+x}-\sqrt{a}}{x\sqrt{a^2+ax}}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Rationalizing the numerator, we get}
\displaystyle \lim \limits_{x \to 0}\frac{\sqrt{a+x}-\sqrt{a}}{x\sqrt{a^2+ax}}
\displaystyle =\lim \limits_{x \to 0}\frac{(\sqrt{a+x}-\sqrt{a})(\sqrt{a+x}+\sqrt{a})}{x\sqrt{a^2+ax}(\sqrt{a+x}+\sqrt{a})}
\displaystyle =\lim \limits_{x \to 0}\frac{a+x-a}{x\sqrt{a^2+ax}(\sqrt{a+x}+\sqrt{a})}
\displaystyle =\lim \limits_{x \to 0}\frac{x}{x\sqrt{a^2+ax}(\sqrt{a+x}+\sqrt{a})}
\displaystyle =\lim \limits_{x \to 0}\frac{1}{\sqrt{a^2+ax}(\sqrt{a+x}+\sqrt{a})}
\displaystyle =\frac{1}{\sqrt{a^2}(\sqrt{a}+\sqrt{a})}
\displaystyle =\frac{1}{a(2\sqrt{a})}
\displaystyle =\frac{1}{2a\sqrt{a}}
\displaystyle \\

\displaystyle \textbf{Question 17: } \lim \limits_{x \to 5}\frac{x-5}{\sqrt{6x-5}-\sqrt{4x+5}}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=5,\text{ the expression }\frac{x-5}{\sqrt{6x-5}-\sqrt{4x+5}}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Rationalizing the denominator, we get}
\displaystyle \lim \limits_{x \to 5}\frac{x-5}{\sqrt{6x-5}-\sqrt{4x+5}}
\displaystyle =\lim \limits_{x \to 5}\frac{(x-5)(\sqrt{6x-5}+\sqrt{4x+5})}{(\sqrt{6x-5}-\sqrt{4x+5})(\sqrt{6x-5}+\sqrt{4x+5})}
\displaystyle =\lim \limits_{x \to 5}\frac{(x-5)(\sqrt{6x-5}+\sqrt{4x+5})}{(6x-5)-(4x+5)}
\displaystyle =\lim \limits_{x \to 5}\frac{(x-5)(\sqrt{6x-5}+\sqrt{4x+5})}{2x-10}
\displaystyle =\lim \limits_{x \to 5}\frac{(x-5)(\sqrt{6x-5}+\sqrt{4x+5})}{2(x-5)}
\displaystyle =\lim \limits_{x \to 5}\frac{\sqrt{6x-5}+\sqrt{4x+5}}{2}
\displaystyle =\frac{\sqrt{6(5)-5}+\sqrt{4(5)+5}}{2}
\displaystyle =\frac{\sqrt{25}+\sqrt{25}}{2}
\displaystyle =\frac{5+5}{2}
\displaystyle =5
\displaystyle \\

\displaystyle \textbf{Question 18: } \lim \limits_{x \to 1}\frac{\sqrt{5x-4}-\sqrt{x}}{x^3-1}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=1,\text{ the expression }\frac{\sqrt{5x-4}-\sqrt{x}}{x^3-1}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Rationalizing the numerator, we get}
\displaystyle \lim \limits_{x \to 1}\frac{\sqrt{5x-4}-\sqrt{x}}{x^3-1}
\displaystyle =\lim \limits_{x \to 1}\frac{(\sqrt{5x-4}-\sqrt{x})(\sqrt{5x-4}+\sqrt{x})}{(x^3-1)(\sqrt{5x-4}+\sqrt{x})}
\displaystyle =\lim \limits_{x \to 1}\frac{5x-4-x}{(x-1)(x^2+x+1)(\sqrt{5x-4}+\sqrt{x})}
\displaystyle =\lim \limits_{x \to 1}\frac{4(x-1)}{(x-1)(x^2+x+1)(\sqrt{5x-4}+\sqrt{x})}
\displaystyle =\lim \limits_{x \to 1}\frac{4}{(x^2+x+1)(\sqrt{5x-4}+\sqrt{x})}
\displaystyle =\frac{4}{(1+1+1)(\sqrt{1}+\sqrt{1})}
\displaystyle =\frac{4}{3\times2}
\displaystyle =\frac{2}{3}
\displaystyle \\

\displaystyle \textbf{Question 19: } \lim \limits_{x \to 2}\frac{\sqrt{1+4x}-\sqrt{5+2x}}{x-2}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=2,\text{ the expression }\frac{\sqrt{1+4x}-\sqrt{5+2x}}{x-2}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Rationalizing the numerator, we get}
\displaystyle \lim \limits_{x \to 2}\frac{\sqrt{1+4x}-\sqrt{5+2x}}{x-2}
\displaystyle =\lim \limits_{x \to 2}\frac{(\sqrt{1+4x}-\sqrt{5+2x})(\sqrt{1+4x}+\sqrt{5+2x})}{(x-2)(\sqrt{1+4x}+\sqrt{5+2x})}
\displaystyle =\lim \limits_{x \to 2}\frac{(1+4x)-(5+2x)}{(x-2)(\sqrt{1+4x}+\sqrt{5+2x})}
\displaystyle =\lim \limits_{x \to 2}\frac{2(x-2)}{(x-2)(\sqrt{1+4x}+\sqrt{5+2x})}
\displaystyle =\lim \limits_{x \to 2}\frac{2}{\sqrt{1+4x}+\sqrt{5+2x}}
\displaystyle =\frac{2}{\sqrt{1+4(2)}+\sqrt{5+2(2)}}
\displaystyle =\frac{2}{3+3}
\displaystyle =\frac{1}{3}
\displaystyle \\

\displaystyle \textbf{Question 20: } \lim \limits_{x \to 1}\frac{\sqrt{3+x}-\sqrt{5-x}}{x^2-1}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=1,\text{ the expression }\frac{\sqrt{3+x}-\sqrt{5-x}}{x^2-1}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Rationalizing the numerator, we get}
\displaystyle \lim \limits_{x \to 1}\frac{\sqrt{3+x}-\sqrt{5-x}}{x^2-1}
\displaystyle =\lim \limits_{x \to 1}\frac{(\sqrt{3+x}-\sqrt{5-x})(\sqrt{3+x}+\sqrt{5-x})}{(x^2-1)(\sqrt{3+x}+\sqrt{5-x})}
\displaystyle =\lim \limits_{x \to 1}\frac{(3+x)-(5-x)}{(x-1)(x+1)(\sqrt{3+x}+\sqrt{5-x})}
\displaystyle =\lim \limits_{x \to 1}\frac{2(x-1)}{(x-1)(x+1)(\sqrt{3+x}+\sqrt{5-x})}
\displaystyle =\lim \limits_{x \to 1}\frac{2}{(x+1)(\sqrt{3+x}+\sqrt{5-x})}
\displaystyle =\frac{2}{(1+1)(\sqrt{3+1}+\sqrt{5-1})}
\displaystyle =\frac{2}{2(2+2)}
\displaystyle =\frac{1}{4}
\displaystyle \\

\displaystyle \textbf{Question 21: } \lim \limits_{x \to 0}\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{x}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=0,\text{ the expression }\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{x}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Rationalizing the numerator, we get}
\displaystyle \lim \limits_{x \to 0}\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{x}
\displaystyle =\lim \limits_{x \to 0}\frac{(\sqrt{1+x^2}-\sqrt{1-x^2})(\sqrt{1+x^2}+\sqrt{1-x^2})}{x(\sqrt{1+x^2}+\sqrt{1-x^2})}
\displaystyle =\lim \limits_{x \to 0}\frac{(1+x^2)-(1-x^2)}{x(\sqrt{1+x^2}+\sqrt{1-x^2})}
\displaystyle =\lim \limits_{x \to 0}\frac{2x^2}{x(\sqrt{1+x^2}+\sqrt{1-x^2})}
\displaystyle =\lim \limits_{x \to 0}\frac{2x}{\sqrt{1+x^2}+\sqrt{1-x^2}}
\displaystyle =\frac{2\times0}{\sqrt{1+0}+\sqrt{1-0}}
\displaystyle =0
\displaystyle \\

\displaystyle \textbf{Question 22: } \lim \limits_{x \to 0}\frac{\sqrt{1+x+x^2}-\sqrt{x+1}}{2x^2}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=0,\text{ the expression }\frac{\sqrt{1+x+x^2}-\sqrt{x+1}}{2x^2}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Rationalizing the numerator, we get}
\displaystyle \lim \limits_{x \to 0}\frac{\sqrt{1+x+x^2}-\sqrt{x+1}}{2x^2}
\displaystyle =\lim \limits_{x \to 0}\frac{(\sqrt{1+x+x^2}-\sqrt{x+1})(\sqrt{1+x+x^2}+\sqrt{x+1})}{2x^2(\sqrt{1+x+x^2}+\sqrt{x+1})}
\displaystyle =\lim \limits_{x \to 0}\frac{(1+x+x^2)-(x+1)}{2x^2(\sqrt{1+x+x^2}+\sqrt{x+1})}
\displaystyle =\lim \limits_{x \to 0}\frac{x^2}{2x^2(\sqrt{1+x+x^2}+\sqrt{x+1})}
\displaystyle =\lim \limits_{x \to 0}\frac{1}{2(\sqrt{1+x+x^2}+\sqrt{x+1})}
\displaystyle =\frac{1}{2(\sqrt{1+0+0^2}+\sqrt{0+1})}
\displaystyle =\frac{1}{2(1+1)}
\displaystyle =\frac{1}{4}
\displaystyle \\

\displaystyle \textbf{Question 23: } \lim \limits_{x \to 4}\frac{2-\sqrt{x}}{4-x}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=4,\text{ the expression }\frac{2-\sqrt{x}}{4-x}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Rationalizing the numerator, we get}
\displaystyle \lim \limits_{x \to 4}\frac{2-\sqrt{x}}{4-x}
\displaystyle =\lim \limits_{x \to 4}\frac{2-\sqrt{x}}{2^2-(\sqrt{x})^2}
\displaystyle =\lim \limits_{x \to 4}\frac{2-\sqrt{x}}{(2-\sqrt{x})(2+\sqrt{x})}
\displaystyle =\lim \limits_{x \to 4}\frac{1}{2+\sqrt{x}}
\displaystyle =\frac{1}{2+\sqrt{4}}
\displaystyle =\frac{1}{2+2}
\displaystyle =\frac{1}{4}
\displaystyle \\

\displaystyle \textbf{Question 24: } \lim \limits_{x \to a}\frac{x-a}{\sqrt{x}-\sqrt{a}},\quad a>0.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=a,\text{ the expression }\frac{x-a}{\sqrt{x}-\sqrt{a}}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Rationalizing the denominator, we get}
\displaystyle \lim \limits_{x \to a}\frac{x-a}{\sqrt{x}-\sqrt{a}}
\displaystyle =\lim \limits_{x \to a}\frac{(\sqrt{x})^2-(\sqrt{a})^2}{\sqrt{x}-\sqrt{a}}
\displaystyle =\lim \limits_{x \to a}\frac{(\sqrt{x}-\sqrt{a})(\sqrt{x}+\sqrt{a})}{\sqrt{x}-\sqrt{a}}
\displaystyle =\lim \limits_{x \to a}\left(\sqrt{x}+\sqrt{a}\right)
\displaystyle =\sqrt{a}+\sqrt{a}
\displaystyle =2\sqrt{a}
\displaystyle \\

\displaystyle \textbf{Question 25: } \lim \limits_{x \to 0}\frac{\sqrt{1+3x}-\sqrt{1-3x}}{x}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=0,\text{ the expression }\frac{\sqrt{1+3x}-\sqrt{1-3x}}{x}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Rationalizing the numerator, we get}
\displaystyle \lim \limits_{x \to 0}\frac{\sqrt{1+3x}-\sqrt{1-3x}}{x}
\displaystyle =\lim \limits_{x \to 0}\frac{(\sqrt{1+3x}-\sqrt{1-3x})(\sqrt{1+3x}+\sqrt{1-3x})}{x(\sqrt{1+3x}+\sqrt{1-3x})}
\displaystyle =\lim \limits_{x \to 0}\frac{(1+3x)-(1-3x)}{x(\sqrt{1+3x}+\sqrt{1-3x})}
\displaystyle =\lim \limits_{x \to 0}\frac{6x}{x(\sqrt{1+3x}+\sqrt{1-3x})}
\displaystyle =\lim \limits_{x \to 0}\frac{6}{\sqrt{1+3x}+\sqrt{1-3x}}
\displaystyle =\frac{6}{\sqrt{1}+\sqrt{1}}
\displaystyle =\frac{6}{2}
\displaystyle =3
\displaystyle \\

\displaystyle \textbf{Question 26: } \lim \limits_{x \to 0}\frac{\sqrt{2-x}-\sqrt{2+x}}{x}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=0,\text{ the expression }\frac{\sqrt{2-x}-\sqrt{2+x}}{x}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Rationalizing the numerator, we get}
\displaystyle \lim \limits_{x \to 0}\frac{\sqrt{2-x}-\sqrt{2+x}}{x}
\displaystyle =\lim \limits_{x \to 0}\frac{(\sqrt{2-x}-\sqrt{2+x})(\sqrt{2-x}+\sqrt{2+x})}{x(\sqrt{2-x}+\sqrt{2+x})}
\displaystyle =\lim \limits_{x \to 0}\frac{(2-x)-(2+x)}{x(\sqrt{2-x}+\sqrt{2+x})}
\displaystyle =\lim \limits_{x \to 0}\frac{-2x}{x(\sqrt{2-x}+\sqrt{2+x})}
\displaystyle =\lim \limits_{x \to 0}\frac{-2}{\sqrt{2-x}+\sqrt{2+x}}
\displaystyle =\frac{-2}{\sqrt{2}+\sqrt{2}}
\displaystyle =\frac{-2}{2\sqrt{2}}
\displaystyle =-\frac{1}{\sqrt{2}}
\displaystyle \\

\displaystyle \textbf{Question 27: } \lim \limits_{x \to 1}\frac{\sqrt{3+x}-\sqrt{5-x}}{x^2-1}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=1,\text{ the expression }\frac{\sqrt{3+x}-\sqrt{5-x}}{x^2-1}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Rationalizing the numerator, we get}
\displaystyle \lim \limits_{x \to 1}\frac{\sqrt{3+x}-\sqrt{5-x}}{x^2-1}
\displaystyle =\lim \limits_{x \to 1}\frac{(\sqrt{3+x}-\sqrt{5-x})(\sqrt{3+x}+\sqrt{5-x})}{(x^2-1)(\sqrt{3+x}+\sqrt{5-x})}
\displaystyle =\lim \limits_{x \to 1}\frac{(3+x)-(5-x)}{(x-1)(x+1)(\sqrt{3+x}+\sqrt{5-x})}
\displaystyle =\lim \limits_{x \to 1}\frac{2(x-1)}{(x-1)(x+1)(\sqrt{3+x}+\sqrt{5-x})}
\displaystyle =\lim \limits_{x \to 1}\frac{2}{(x+1)(\sqrt{3+x}+\sqrt{5-x})}
\displaystyle =\frac{2}{(1+1)(\sqrt{3+1}+\sqrt{5-1})}
\displaystyle =\frac{2}{2(2+2)}
\displaystyle =\frac{2}{8}
\displaystyle =\frac{1}{4}
\displaystyle \\

\displaystyle \textbf{Question 28: } \lim \limits_{x \to 1}\frac{(2x-3)(\sqrt{x}-1)}{3x^2+3x-6}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=1,\text{ the expression }\frac{(2x-3)(\sqrt{x}-1)}{3x^2+3x-6}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Factorizing the denominator, we get}
\displaystyle \lim \limits_{x \to 1}\frac{(2x-3)(\sqrt{x}-1)}{3x^2+3x-6}
\displaystyle =\lim \limits_{x \to 1}\frac{(2x-3)(\sqrt{x}-1)}{3(x^2+x-2)}
\displaystyle =\lim \limits_{x \to 1}\frac{(2x-3)(\sqrt{x}-1)}{3(x-1)(x+2)}
\displaystyle =\lim \limits_{x \to 1}\frac{(2x-3)(\sqrt{x}-1)}{3(\sqrt{x}-1)(\sqrt{x}+1)(x+2)}
\displaystyle =\lim \limits_{x \to 1}\frac{2x-3}{3(\sqrt{x}+1)(x+2)}
\displaystyle =\frac{2(1)-3}{3(\sqrt{1}+1)(1+2)}
\displaystyle =\frac{-1}{3(2)(3)}
\displaystyle =-\frac{1}{18}
\displaystyle \\

\displaystyle \textbf{Question 29: } \lim \limits_{x \to 0}\frac{\sqrt{1+x^2}-\sqrt{1+x}}{\sqrt{1+x^3}-\sqrt{1+x}}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=0,\text{ the expression }\frac{\sqrt{1+x^2}-\sqrt{1+x}}{\sqrt{1+x^3}-\sqrt{1+x}}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Rationalizing the numerator and denominator, we get}
\displaystyle \lim \limits_{x \to 0}\frac{\sqrt{1+x^2}-\sqrt{1+x}}{\sqrt{1+x^3}-\sqrt{1+x}}
\displaystyle =\lim \limits_{x \to 0}\left(\frac{(\sqrt{1+x^2}-\sqrt{1+x})(\sqrt{1+x^2}+\sqrt{1+x})}{\sqrt{1+x^2}+\sqrt{1+x}}\times\frac{\sqrt{1+x^3}+\sqrt{1+x}}{(\sqrt{1+x^3}-\sqrt{1+x})(\sqrt{1+x^3}+\sqrt{1+x})}\right)
\displaystyle =\lim \limits_{x \to 0}\left(\frac{(1+x^2)-(1+x)}{(1+x^3)-(1+x)}\times\frac{\sqrt{1+x^3}+\sqrt{1+x}}{\sqrt{1+x^2}+\sqrt{1+x}}\right)
\displaystyle =\lim \limits_{x \to 0}\left(\frac{x^2-x}{x^3-x}\times\frac{\sqrt{1+x^3}+\sqrt{1+x}}{\sqrt{1+x^2}+\sqrt{1+x}}\right)
\displaystyle =\lim \limits_{x \to 0}\left(\frac{x(x-1)}{x(x-1)(x+1)}\times\frac{\sqrt{1+x^3}+\sqrt{1+x}}{\sqrt{1+x^2}+\sqrt{1+x}}\right)
\displaystyle =\lim \limits_{x \to 0}\left(\frac{1}{x+1}\times\frac{\sqrt{1+x^3}+\sqrt{1+x}}{\sqrt{1+x^2}+\sqrt{1+x}}\right)
\displaystyle =\frac{1}{0+1}\times\frac{\sqrt{1+0}+\sqrt{1+0}}{\sqrt{1+0}+\sqrt{1+0}}
\displaystyle =1\times\frac{2}{2}
\displaystyle =1
\displaystyle \\

\displaystyle \textbf{Question 30: } \lim \limits_{x \to 1}\frac{x^2-\sqrt{x}}{\sqrt{x}-1}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=1,\text{ the expression }\frac{x^2-\sqrt{x}}{\sqrt{x}-1}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Factorizing the numerator, we get}
\displaystyle \lim \limits_{x \to 1}\frac{x^2-\sqrt{x}}{\sqrt{x}-1}
\displaystyle =\lim \limits_{x \to 1}\frac{\sqrt{x}(x\sqrt{x}-1)}{\sqrt{x}-1}
\displaystyle =\lim \limits_{x \to 1}\frac{\sqrt{x}\left((\sqrt{x})^3-1\right)}{\sqrt{x}-1}
\displaystyle =\lim \limits_{x \to 1}\frac{\sqrt{x}(\sqrt{x}-1)(x+\sqrt{x}+1)}{\sqrt{x}-1}
\displaystyle =\lim \limits_{x \to 1}\sqrt{x}(x+\sqrt{x}+1)
\displaystyle =1(1+1+1)
\displaystyle =3
\displaystyle \\

\displaystyle \textbf{Question 31: } \lim \limits_{h \to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h},\quad x\neq 0.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }h=0,\text{ the expression }\frac{\sqrt{x+h}-\sqrt{x}}{h}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Rationalizing the numerator, we get}
\displaystyle \lim \limits_{h \to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}
\displaystyle =\lim \limits_{h \to 0}\frac{(\sqrt{x+h}-\sqrt{x})(\sqrt{x+h}+\sqrt{x})}{h(\sqrt{x+h}+\sqrt{x})}
\displaystyle =\lim \limits_{h \to 0}\frac{(x+h)-x}{h(\sqrt{x+h}+\sqrt{x})}
\displaystyle =\lim \limits_{h \to 0}\frac{h}{h(\sqrt{x+h}+\sqrt{x})}
\displaystyle =\lim \limits_{h \to 0}\frac{1}{\sqrt{x+h}+\sqrt{x}}
\displaystyle =\frac{1}{\sqrt{x}+\sqrt{x}}
\displaystyle =\frac{1}{2\sqrt{x}}
\displaystyle \\

\displaystyle \textbf{Question 32: } \lim \limits_{x \to \sqrt{10}}\frac{\sqrt{7+2x}-(\sqrt{5}+\sqrt{2})}{x^2-10}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=\sqrt{10},\text{ we have }\sqrt{7+2\sqrt{10}}=\sqrt{5}+\sqrt{2}.
\displaystyle \text{Therefore, the expression }\frac{\sqrt{7+2x}-(\sqrt{5}+\sqrt{2})}{x^2-10}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Rationalizing the numerator, we get}
\displaystyle \lim \limits_{x \to \sqrt{10}}\frac{\sqrt{7+2x}-(\sqrt{5}+\sqrt{2})}{x^2-10}
\displaystyle =\lim \limits_{x \to \sqrt{10}}\frac{\sqrt{7+2x}-\sqrt{7+2\sqrt{10}}}{(x-\sqrt{10})(x+\sqrt{10})}
\displaystyle =\lim \limits_{x \to \sqrt{10}}\frac{\left(\sqrt{7+2x}-\sqrt{7+2\sqrt{10}}\right)\left(\sqrt{7+2x}+\sqrt{7+2\sqrt{10}}\right)}{(x-\sqrt{10})(x+\sqrt{10})\left(\sqrt{7+2x}+\sqrt{7+2\sqrt{10}}\right)}
\displaystyle =\lim \limits_{x \to \sqrt{10}}\frac{(7+2x)-(7+2\sqrt{10})}{(x-\sqrt{10})(x+\sqrt{10})\left(\sqrt{7+2x}+\sqrt{7+2\sqrt{10}}\right)}
\displaystyle =\lim \limits_{x \to \sqrt{10}}\frac{2(x-\sqrt{10})}{(x-\sqrt{10})(x+\sqrt{10})\left(\sqrt{7+2x}+\sqrt{7+2\sqrt{10}}\right)}
\displaystyle =\lim \limits_{x \to \sqrt{10}}\frac{2}{(x+\sqrt{10})\left(\sqrt{7+2x}+\sqrt{7+2\sqrt{10}}\right)}
\displaystyle =\frac{2}{(2\sqrt{10})\left(2\sqrt{7+2\sqrt{10}}\right)}
\displaystyle =\frac{1}{2\sqrt{10}\sqrt{7+2\sqrt{10}}}
\displaystyle =\frac{1}{2\sqrt{10}(\sqrt{5}+\sqrt{2})}
\displaystyle =\frac{\sqrt{5}-\sqrt{2}}{2\sqrt{10}(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})}
\displaystyle =\frac{\sqrt{5}-\sqrt{2}}{6\sqrt{10}}
\displaystyle \\

\displaystyle \textbf{Question 33: } \lim \limits_{x \to \sqrt{6}}\frac{\sqrt{5+2x}-(\sqrt{3}+\sqrt{2})}{x^2-6}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=\sqrt{6},\text{ the expression }\frac{\sqrt{5+2x}-(\sqrt{3}+\sqrt{2})}{x^2-6}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Since }(\sqrt{3}+\sqrt{2})^2=5+2\sqrt{6},\text{ we have }\sqrt{5+2\sqrt{6}}=\sqrt{3}+\sqrt{2}.
\displaystyle \text{Rationalizing the numerator, we get}
\displaystyle \lim \limits_{x \to \sqrt{6}}\frac{\sqrt{5+2x}-(\sqrt{3}+\sqrt{2})}{x^2-6}
\displaystyle =\lim \limits_{x \to \sqrt{6}}\frac{\sqrt{5+2x}-\sqrt{5+2\sqrt{6}}}{(x-\sqrt{6})(x+\sqrt{6})}
\displaystyle =\lim \limits_{x \to \sqrt{6}}\frac{(\sqrt{5+2x}-\sqrt{5+2\sqrt{6}})(\sqrt{5+2x}+\sqrt{5+2\sqrt{6}})}{(x-\sqrt{6})(x+\sqrt{6})(\sqrt{5+2x}+\sqrt{5+2\sqrt{6}})}
\displaystyle =\lim \limits_{x \to \sqrt{6}}\frac{2(x-\sqrt{6})}{(x-\sqrt{6})(x+\sqrt{6})(\sqrt{5+2x}+\sqrt{5+2\sqrt{6}})}
\displaystyle =\lim \limits_{x \to \sqrt{6}}\frac{2}{(x+\sqrt{6})(\sqrt{5+2x}+\sqrt{5+2\sqrt{6}})}
\displaystyle =\frac{1}{2\sqrt{6}(\sqrt{3}+\sqrt{2})}
\displaystyle =\frac{\sqrt{3}-\sqrt{2}}{2\sqrt{6}(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}
\displaystyle =\frac{\sqrt{3}-\sqrt{2}}{2\sqrt{6}(3-2)}
\displaystyle =\frac{\sqrt{3}-\sqrt{2}}{2\sqrt{6}}
\displaystyle \\

\displaystyle \textbf{Question 34: } \lim \limits_{x \to \sqrt{2}}\frac{\sqrt{3+2x}-(\sqrt{2}+1)}{x^2-2}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=\sqrt{2},\text{ the expression }\frac{\sqrt{3+2x}-(\sqrt{2}+1)}{x^2-2}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Since }(\sqrt{2}+1)^2=3+2\sqrt{2},\text{ we have }\sqrt{3+2\sqrt{2}}=\sqrt{2}+1.
\displaystyle \text{Rationalizing the numerator, we get}
\displaystyle \lim \limits_{x \to \sqrt{2}}\frac{\sqrt{3+2x}-(\sqrt{2}+1)}{x^2-2}
\displaystyle =\lim \limits_{x \to \sqrt{2}}\frac{\sqrt{3+2x}-\sqrt{3+2\sqrt{2}}}{(x-\sqrt{2})(x+\sqrt{2})}
\displaystyle =\lim \limits_{x \to \sqrt{2}}\frac{(\sqrt{3+2x}-\sqrt{3+2\sqrt{2}})(\sqrt{3+2x}+\sqrt{3+2\sqrt{2}})}{(x-\sqrt{2})(x+\sqrt{2})(\sqrt{3+2x}+\sqrt{3+2\sqrt{2}})}
\displaystyle =\lim \limits_{x \to \sqrt{2}}\frac{(3+2x)-(3+2\sqrt{2})}{(x-\sqrt{2})(x+\sqrt{2})(\sqrt{3+2x}+\sqrt{3+2\sqrt{2}})}
\displaystyle =\lim \limits_{x \to \sqrt{2}}\frac{2(x-\sqrt{2})}{(x-\sqrt{2})(x+\sqrt{2})(\sqrt{3+2x}+\sqrt{3+2\sqrt{2}})}
\displaystyle =\lim \limits_{x \to \sqrt{2}}\frac{2}{(x+\sqrt{2})(\sqrt{3+2x}+\sqrt{3+2\sqrt{2}})}
\displaystyle =\frac{2}{(2\sqrt{2})(2\sqrt{3+2\sqrt{2}})}
\displaystyle =\frac{1}{2\sqrt{2}(\sqrt{2}+1)}
\displaystyle =\frac{\sqrt{2}-1}{2\sqrt{2}(\sqrt{2}+1)(\sqrt{2}-1)}
\displaystyle =\frac{\sqrt{2}-1}{2\sqrt{2}}
\displaystyle \\

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