\displaystyle \textbf{Question 1: } \lim \limits_{x \to -5}\frac{2x^2+9x-5}{x+5}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=-5,\text{ the expression }\frac{2x^2+9x-5}{x+5}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Therefore, }(x+5)\text{ is a common factor of the numerator and denominator.}
\displaystyle \lim \limits_{x \to -5}\frac{2x^2+9x-5}{x+5}
\displaystyle =\lim \limits_{x \to -5}\frac{2x^2+10x-x-5}{x+5}
\displaystyle =\lim \limits_{x \to -5}\frac{2x(x+5)-(x+5)}{x+5}
\displaystyle =\lim \limits_{x \to -5}\frac{(2x-1)(x+5)}{x+5}
\displaystyle =\lim \limits_{x \to -5}(2x-1)
\displaystyle =2(-5)-1
\displaystyle =-11
\displaystyle \\

\displaystyle \textbf{Question 2: } \lim \limits_{x \to 3}\frac{x^2-4x+3}{x^2-2x-3}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=3,\text{ the expression }\frac{x^2-4x+3}{x^2-2x-3}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Therefore, }(x-3)\text{ is a common factor of the numerator and denominator.}
\displaystyle \lim \limits_{x \to 3}\frac{x^2-4x+3}{x^2-2x-3}
\displaystyle =\lim \limits_{x \to 3}\frac{x^2-x-3x+3}{x^2-3x+x-3}
\displaystyle =\lim \limits_{x \to 3}\frac{(x-3)(x-1)}{(x+1)(x-3)}
\displaystyle =\lim \limits_{x \to 3}\frac{x-1}{x+1}
\displaystyle =\frac{3-1}{3+1}
\displaystyle =\frac{1}{2}
\displaystyle \\

\displaystyle \textbf{Question 3: } \lim \limits_{x \to 3}\frac{x^4-81}{x^2-9}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=3,\text{ the expression }\frac{x^4-81}{x^2-9}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Therefore, }(x-3)\text{ is a factor of both the numerator and denominator.}
\displaystyle \lim \limits_{x \to 3}\frac{x^4-81}{x^2-9}
\displaystyle =\lim \limits_{x \to 3}\frac{(x^2)^2-9^2}{x^2-9}
\displaystyle =\lim \limits_{x \to 3}\frac{(x^2-9)(x^2+9)}{x^2-9}
\displaystyle =\lim \limits_{x \to 3}(x^2+9)
\displaystyle =3^2+9
\displaystyle =18
\displaystyle \\

\displaystyle \textbf{Question 4: } \lim \limits_{x \to 2}\frac{x^3-8}{x^2-4}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=2,\text{ the expression }\frac{x^3-8}{x^2-4}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Therefore, }(x-2)\text{ is a common factor of the numerator and denominator.}
\displaystyle \lim \limits_{x \to 2}\frac{x^3-8}{x^2-4}
\displaystyle =\lim \limits_{x \to 2}\frac{(x-2)(x^2+2x+4)}{(x-2)(x+2)}
\displaystyle =\lim \limits_{x \to 2}\frac{x^2+2x+4}{x+2}
\displaystyle =\frac{2^2+2(2)+4}{2+2}
\displaystyle =\frac{12}{4}
\displaystyle =3
\displaystyle \\

\displaystyle \textbf{Question 5: } \lim \limits_{x \to-\frac{1}{2}}\frac{8x^3+1}{2x+1}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=-\frac{1}{2},\text{ the expression }\frac{8x^3+1}{2x+1}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Therefore, }(2x+1)\text{ is a common factor of the numerator and denominator.}
\displaystyle \lim \limits_{x \to-\frac{1}{2}}\frac{8x^3+1}{2x+1}
\displaystyle =\lim \limits_{x \to-\frac{1}{2}}\frac{(2x)^3+1^3}{2x+1}
\displaystyle =\lim \limits_{x \to-\frac{1}{2}}\frac{(2x+1)(4x^2-2x+1)}{2x+1}
\displaystyle =\lim \limits_{x \to-\frac{1}{2}}(4x^2-2x+1)
\displaystyle =4\left(-\frac{1}{2}\right)^2-2\left(-\frac{1}{2}\right)+1
\displaystyle =1+1+1
\displaystyle =3
\displaystyle \\

\displaystyle \textbf{Question 6: } \lim \limits_{x \to 4}\frac{x^2-7x+12}{x^2-3x-4}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=4,\text{ the expression }\frac{x^2-7x+12}{x^2-3x-4}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Therefore, }(x-4)\text{ is a common factor of the numerator and denominator.}
\displaystyle \lim \limits_{x \to 4}\frac{x^2-7x+12}{x^2-3x-4}
\displaystyle =\lim \limits_{x \to 4}\frac{x^2-3x-4x+12}{x^2-4x+x-4}
\displaystyle =\lim \limits_{x \to 4}\frac{x(x-3)-4(x-3)}{(x-4)(x+1)}
\displaystyle =\lim \limits_{x \to 4}\frac{(x-4)(x-3)}{(x-4)(x+1)}
\displaystyle =\lim \limits_{x \to 4}\frac{x-3}{x+1}
\displaystyle =\frac{4-3}{4+1}
\displaystyle =\frac{1}{5}
\displaystyle \\

\displaystyle \textbf{Question 7: } \lim \limits_{x \to 2}\frac{x^4-16}{x-2}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=2,\text{ the expression }\frac{x^4-16}{x-2}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Therefore, }(x-2)\text{ is a common factor of the numerator and denominator.}
\displaystyle \lim \limits_{x \to 2}\frac{x^4-16}{x-2}
\displaystyle =\lim \limits_{x \to 2}\frac{(x^2)^2-4^2}{x-2}
\displaystyle =\lim \limits_{x \to 2}\frac{(x^2-4)(x^2+4)}{x-2}
\displaystyle =\lim \limits_{x \to 2}\frac{(x-2)(x+2)(x^2+4)}{x-2}
\displaystyle =\lim \limits_{x \to 2}(x+2)(x^2+4)
\displaystyle =(2+2)(2^2+4)
\displaystyle =4\times8
\displaystyle =32
\displaystyle \\

\displaystyle \textbf{Question 8: } \lim \limits_{x \to 5}\frac{x^2-9x+20}{x^2-6x+5}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=5,\text{ the expression }\frac{x^2-9x+20}{x^2-6x+5}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Therefore, }(x-5)\text{ is a common factor of the numerator and denominator.}
\displaystyle \lim \limits_{x \to 5}\frac{x^2-9x+20}{x^2-6x+5}
\displaystyle =\lim \limits_{x \to 5}\frac{x^2-4x-5x+20}{x^2-x-5x+5}
\displaystyle =\lim \limits_{x \to 5}\frac{x(x-4)-5(x-4)}{x(x-1)-5(x-1)}
\displaystyle =\lim \limits_{x \to 5}\frac{(x-5)(x-4)}{(x-1)(x-5)}
\displaystyle =\lim \limits_{x \to 5}\frac{x-4}{x-1}
\displaystyle =\frac{5-4}{5-1}
\displaystyle =\frac{1}{4}
\displaystyle \\

\displaystyle \textbf{Question 9: } \lim \limits_{x \to -1}\frac{x^3+1}{x+1}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=-1,\text{ the expression }\frac{x^3+1}{x+1}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Therefore, }(x+1)\text{ is a common factor of the numerator and denominator.}
\displaystyle \lim \limits_{x \to -1}\frac{x^3+1}{x+1}
\displaystyle =\lim \limits_{x \to -1}\frac{x^3+1^3}{x+1}
\displaystyle =\lim \limits_{x \to -1}\frac{(x+1)(x^2-x+1)}{x+1}
\displaystyle =\lim \limits_{x \to -1}(x^2-x+1)
\displaystyle =(-1)^2-(-1)+1
\displaystyle =1+1+1
\displaystyle =3
\displaystyle \\

\displaystyle \textbf{Question 10: } \lim \limits_{x \to 5}\frac{x^3-125}{x^2-7x+10}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=5,\text{ the expression }\frac{x^3-125}{x^2-7x+10}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Therefore, }(x-5)\text{ is a common factor of the numerator and denominator.}
\displaystyle \lim \limits_{x \to 5}\frac{x^3-125}{x^2-7x+10}
\displaystyle =\lim \limits_{x \to 5}\frac{x^3-5^3}{x^2-2x-5x+10}
\displaystyle =\lim \limits_{x \to 5}\frac{(x-5)(x^2+5x+25)}{x(x-2)-5(x-2)}
\displaystyle =\lim \limits_{x \to 5}\frac{(x-5)(x^2+5x+25)}{(x-5)(x-2)}
\displaystyle =\lim \limits_{x \to 5}\frac{x^2+5x+25}{x-2}
\displaystyle =\frac{5^2+5(5)+25}{5-2}
\displaystyle =\frac{75}{3}
\displaystyle =25
\displaystyle \\

\displaystyle \textbf{Question 11: } \lim \limits_{x \to \sqrt{2}}\frac{x^2-2}{x^2+\sqrt{2}x-4}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=\sqrt{2},\text{ the expression }\frac{x^2-2}{x^2+\sqrt{2}x-4}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Therefore, }(x-\sqrt{2})\text{ is a common factor of the numerator and denominator.}
\displaystyle \lim \limits_{x \to \sqrt{2}}\frac{x^2-2}{x^2+\sqrt{2}x-4}
\displaystyle =\lim \limits_{x \to \sqrt{2}}\frac{x^2-(\sqrt{2})^2}{x^2+2\sqrt{2}x-\sqrt{2}x-4}
\displaystyle =\lim \limits_{x \to \sqrt{2}}\frac{(x-\sqrt{2})(x+\sqrt{2})}{x(x+2\sqrt{2})-\sqrt{2}(x+2\sqrt{2})}
\displaystyle =\lim \limits_{x \to \sqrt{2}}\frac{(x-\sqrt{2})(x+\sqrt{2})}{(x-\sqrt{2})(x+2\sqrt{2})}
\displaystyle =\lim \limits_{x \to \sqrt{2}}\frac{x+\sqrt{2}}{x+2\sqrt{2}}
\displaystyle =\frac{\sqrt{2}+\sqrt{2}}{\sqrt{2}+2\sqrt{2}}
\displaystyle =\frac{2\sqrt{2}}{3\sqrt{2}}
\displaystyle =\frac{2}{3}
\displaystyle \\

\displaystyle \textbf{Question 12: } \lim \limits_{x \to \sqrt{3}}\frac{x^2-3}{x^2+3\sqrt{3}x-12}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=\sqrt{3},\text{ the expression }\frac{x^2-3}{x^2+3\sqrt{3}x-12}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Therefore, }(x-\sqrt{3})\text{ is a common factor of the numerator and denominator.}
\displaystyle \lim \limits_{x \to \sqrt{3}}\frac{x^2-3}{x^2+3\sqrt{3}x-12}
\displaystyle =\lim \limits_{x \to \sqrt{3}}\frac{x^2-(\sqrt{3})^2}{x^2+4\sqrt{3}x-\sqrt{3}x-12}
\displaystyle =\lim \limits_{x \to \sqrt{3}}\frac{(x-\sqrt{3})(x+\sqrt{3})}{x(x+4\sqrt{3})-\sqrt{3}(x+4\sqrt{3})}
\displaystyle =\lim \limits_{x \to \sqrt{3}}\frac{(x-\sqrt{3})(x+\sqrt{3})}{(x-\sqrt{3})(x+4\sqrt{3})}
\displaystyle =\lim \limits_{x \to \sqrt{3}}\frac{x+\sqrt{3}}{x+4\sqrt{3}}
\displaystyle =\frac{\sqrt{3}+\sqrt{3}}{\sqrt{3}+4\sqrt{3}}
\displaystyle =\frac{2\sqrt{3}}{5\sqrt{3}}
\displaystyle =\frac{2}{5}
\displaystyle \\

\displaystyle \textbf{Question 13: } \lim \limits_{x \to \sqrt{3}}\frac{x^4-9}{x^2+4\sqrt{3}x-15}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=\sqrt{3},\text{ the expression }\frac{x^4-9}{x^2+4\sqrt{3}x-15}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Therefore, }(x-\sqrt{3})\text{ is a common factor of the numerator and denominator.}
\displaystyle \lim \limits_{x \to \sqrt{3}}\frac{x^4-9}{x^2+4\sqrt{3}x-15}
\displaystyle =\lim \limits_{x \to \sqrt{3}}\frac{(x^2)^2-3^2}{x^2+5\sqrt{3}x-\sqrt{3}x-15}
\displaystyle =\lim \limits_{x \to \sqrt{3}}\frac{(x^2-3)(x^2+3)}{x(x+5\sqrt{3})-\sqrt{3}(x+5\sqrt{3})}
\displaystyle =\lim \limits_{x \to \sqrt{3}}\frac{(x^2-3)(x^2+3)}{(x-\sqrt{3})(x+5\sqrt{3})}
\displaystyle =\lim \limits_{x \to \sqrt{3}}\frac{(x-\sqrt{3})(x+\sqrt{3})(x^2+3)}{(x-\sqrt{3})(x+5\sqrt{3})}
\displaystyle =\lim \limits_{x \to \sqrt{3}}\frac{(x+\sqrt{3})(x^2+3)}{x+5\sqrt{3}}
\displaystyle =\frac{(\sqrt{3}+\sqrt{3})(3+3)}{\sqrt{3}+5\sqrt{3}}
\displaystyle =\frac{2\sqrt{3}\times6}{6\sqrt{3}}
\displaystyle =2
\displaystyle \\

\displaystyle \textbf{Question 14: } \lim \limits_{x \to 2}\left(\frac{x}{x-2}-\frac{4}{x^2-2x}\right).
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=2,\text{ the expression assumes the indeterminate form }\infty-\infty.
\displaystyle \text{Combining the two fractions, we get}
\displaystyle \lim \limits_{x \to 2}\left(\frac{x}{x-2}-\frac{4}{x^2-2x}\right)
\displaystyle =\lim \limits_{x \to 2}\left(\frac{x}{x-2}-\frac{4}{x(x-2)}\right)
\displaystyle =\lim \limits_{x \to 2}\frac{x^2-4}{x(x-2)}
\displaystyle =\lim \limits_{x \to 2}\frac{(x-2)(x+2)}{x(x-2)}
\displaystyle =\lim \limits_{x \to 2}\frac{x+2}{x}
\displaystyle =\frac{2+2}{2}
\displaystyle =2
\displaystyle \\

\displaystyle \textbf{Question 15: } \lim \limits_{x \to 1}\left(\frac{1}{x^2+x-2}-\frac{x}{x^3-1}\right).
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=1,\text{ the expression assumes the indeterminate form }\infty-\infty.
\displaystyle \text{Combining the two fractions, we get}
\displaystyle \lim \limits_{x \to 1}\left(\frac{1}{x^2+x-2}-\frac{x}{x^3-1}\right)
\displaystyle =\lim \limits_{x \to 1}\frac{(x^3-1)-x(x^2+x-2)}{(x^2+x-2)(x^3-1)}
\displaystyle =\lim \limits_{x \to 1}\frac{x^3-1-x^3-x^2+2x}{(x^2+x-2)(x^3-1)}
\displaystyle =\lim \limits_{x \to 1}\frac{-x^2+2x-1}{(x^2+x-2)(x^3-1)}
\displaystyle =\lim \limits_{x \to 1}\frac{-(x-1)^2}{(x-1)(x+2)(x-1)(x^2+x+1)}
\displaystyle =\lim \limits_{x \to 1}\frac{-1}{(x+2)(x^2+x+1)}
\displaystyle =\frac{-1}{(1+2)(1+1+1)}
\displaystyle =-\frac{1}{9}
\displaystyle \\

\displaystyle \textbf{Question 16: } \lim \limits_{x \to 3}\left(\frac{1}{x-3}-\frac{2}{x^2-4x+3}\right).
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=3,\text{ the expression assumes the indeterminate form }\infty-\infty.
\displaystyle \text{Factorizing the second denominator and combining the fractions, we get}
\displaystyle \lim \limits_{x \to 3}\left(\frac{1}{x-3}-\frac{2}{x^2-4x+3}\right)
\displaystyle =\lim \limits_{x \to 3}\left(\frac{1}{x-3}-\frac{2}{x^2-3x-x+3}\right)
\displaystyle =\lim \limits_{x \to 3}\left(\frac{1}{x-3}-\frac{2}{x(x-3)-(x-3)}\right)
\displaystyle =\lim \limits_{x \to 3}\left(\frac{1}{x-3}-\frac{2}{(x-3)(x-1)}\right)
\displaystyle =\lim \limits_{x \to 3}\frac{x-1-2}{(x-3)(x-1)}
\displaystyle =\lim \limits_{x \to 3}\frac{x-3}{(x-3)(x-1)}
\displaystyle =\lim \limits_{x \to 3}\frac{1}{x-1}
\displaystyle =\frac{1}{3-1}
\displaystyle =\frac{1}{2}
\displaystyle \\

\displaystyle \textbf{Question 17: } \lim \limits_{x \to 2}\left(\frac{1}{x-2}-\frac{2}{x^2-2x}\right).
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=2,\text{ the expression assumes the indeterminate form }\infty-\infty.
\displaystyle \text{Factorizing the second denominator and combining the fractions, we get}
\displaystyle \lim \limits_{x \to 2}\left(\frac{1}{x-2}-\frac{2}{x^2-2x}\right)
\displaystyle =\lim \limits_{x \to 2}\left(\frac{1}{x-2}-\frac{2}{x(x-2)}\right)
\displaystyle =\lim \limits_{x \to 2}\frac{x-2}{x(x-2)}
\displaystyle =\lim \limits_{x \to 2}\frac{1}{x}
\displaystyle =\frac{1}{2}
\displaystyle \\

\displaystyle \textbf{Question 18: } \lim \limits_{x \to \frac{1}{4}}\frac{4x-1}{2\sqrt{x}-1}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=\frac{1}{4},\text{ the expression }\frac{4x-1}{2\sqrt{x}-1}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Rewriting the numerator as a difference of squares, we get}
\displaystyle \lim \limits_{x \to \frac{1}{4}}\frac{4x-1}{2\sqrt{x}-1}
\displaystyle =\lim \limits_{x \to \frac{1}{4}}\frac{(2\sqrt{x})^2-1^2}{2\sqrt{x}-1}
\displaystyle =\lim \limits_{x \to \frac{1}{4}}\frac{(2\sqrt{x}-1)(2\sqrt{x}+1)}{2\sqrt{x}-1}
\displaystyle =\lim \limits_{x \to \frac{1}{4}}(2\sqrt{x}+1)
\displaystyle =2\sqrt{\frac{1}{4}}+1
\displaystyle =2
\displaystyle \\

\displaystyle \textbf{Question 19: } \lim \limits_{x \to 4}\frac{x^2-16}{\sqrt{x}-2}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=4,\text{ the expression }\frac{x^2-16}{\sqrt{x}-2}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Factorizing the numerator, we get}
\displaystyle \lim \limits_{x \to 4}\frac{x^2-16}{\sqrt{x}-2}
\displaystyle =\lim \limits_{x \to 4}\frac{x^2-4^2}{\sqrt{x}-2}
\displaystyle =\lim \limits_{x \to 4}\frac{(x-4)(x+4)}{\sqrt{x}-2}
\displaystyle =\lim \limits_{x \to 4}\frac{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)}{\sqrt{x}-2}
\displaystyle =\lim \limits_{x \to 4}(\sqrt{x}+2)(x+4)
\displaystyle =(\sqrt{4}+2)(4+4)
\displaystyle =(2+2)(8)
\displaystyle =32
\displaystyle \\

\displaystyle \textbf{Question 20: } \lim \limits_{x \to 0}\frac{(a+x)^2-a^2}{x}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=0,\text{ the expression }\frac{(a+x)^2-a^2}{x}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Therefore, }x\text{ is a common factor of the numerator and denominator.}
\displaystyle \lim \limits_{x \to 0}\frac{(a+x)^2-a^2}{x}
\displaystyle =\lim \limits_{x \to 0}\frac{a^2+x^2+2ax-a^2}{x}
\displaystyle =\lim \limits_{x \to 0}\frac{x(x+2a)}{x}
\displaystyle =\lim \limits_{x \to 0}(x+2a)
\displaystyle =0+2a
\displaystyle =2a
\displaystyle \\

\displaystyle \textbf{Question 21: } \lim \limits_{x \to 2}\left(\frac{1}{x-2}-\frac{4}{x^3-2x^2}\right).
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=2,\text{ the expression assumes the indeterminate form }\infty-\infty.
\displaystyle \text{Writing }x^3-2x^2=x^2(x-2)\text{ and combining the fractions, we get}
\displaystyle \lim \limits_{x \to 2}\left(\frac{1}{x-2}-\frac{4}{x^3-2x^2}\right)
\displaystyle =\lim \limits_{x \to 2}\left(\frac{1}{x-2}-\frac{4}{x^2(x-2)}\right)
\displaystyle =\lim \limits_{x \to 2}\frac{x^2-4}{x^2(x-2)}
\displaystyle =\lim \limits_{x \to 2}\frac{(x-2)(x+2)}{x^2(x-2)}
\displaystyle =\lim \limits_{x \to 2}\frac{x+2}{x^2}
\displaystyle =\frac{2+2}{2^2}
\displaystyle =1
\displaystyle \\

\displaystyle \textbf{Question 22: } \lim \limits_{x \to 3}\left(\frac{1}{x-3}-\frac{3}{x^2-3x}\right).
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=3,\text{ the expression assumes the indeterminate form }\infty-\infty.
\displaystyle \text{Writing }x^2-3x=x(x-3)\text{ and combining the fractions, we get}
\displaystyle \lim \limits_{x \to 3}\left(\frac{1}{x-3}-\frac{3}{x^2-3x}\right)
\displaystyle =\lim \limits_{x \to 3}\left(\frac{1}{x-3}-\frac{3}{x(x-3)}\right)
\displaystyle =\lim \limits_{x \to 3}\frac{x-3}{x(x-3)}
\displaystyle =\lim \limits_{x \to 3}\frac{1}{x}
\displaystyle =\frac{1}{3}
\displaystyle \\

\displaystyle \textbf{Question 23: } \lim \limits_{x \to 1}\left(\frac{1}{x-1}-\frac{2}{x^2-1}\right).
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=1,\text{ the expression assumes the indeterminate form }\infty-\infty.
\displaystyle \text{Factorizing }x^2-1\text{ and combining the fractions, we get}
\displaystyle \lim \limits_{x \to 1}\left(\frac{1}{x-1}-\frac{2}{x^2-1}\right)
\displaystyle =\lim \limits_{x \to 1}\left(\frac{1}{x-1}-\frac{2}{(x-1)(x+1)}\right)
\displaystyle =\lim \limits_{x \to 1}\frac{x+1-2}{(x-1)(x+1)}
\displaystyle =\lim \limits_{x \to 1}\frac{x-1}{(x-1)(x+1)}
\displaystyle =\lim \limits_{x \to 1}\frac{1}{x+1}
\displaystyle =\frac{1}{1+1}
\displaystyle =\frac{1}{2}
\displaystyle \\

\displaystyle \textbf{Question 24: } \lim \limits_{x \to 3}(x^2-9)\left(\frac{1}{x+3}+\frac{1}{x-3}\right).
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=3,\text{ the expression assumes the indeterminate form }0\times\infty.
\displaystyle \text{Combining the fractions and simplifying, we get}
\displaystyle \lim \limits_{x \to 3}(x^2-9)\left(\frac{1}{x+3}+\frac{1}{x-3}\right)
\displaystyle =\lim \limits_{x \to 3}(x^2-9)\left(\frac{x-3+x+3}{(x+3)(x-3)}\right)
\displaystyle =\lim \limits_{x \to 3}(x^2-9)\left(\frac{2x}{x^2-9}\right)
\displaystyle =\lim \limits_{x \to 3}2x
\displaystyle =2(3)
\displaystyle =6
\displaystyle \\

\displaystyle \textbf{Question 25: } \lim \limits_{x \to 1}\frac{x^4-3x^3+2}{x^3-5x^2+3x+1}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=1,\text{ the expression }\frac{x^4-3x^3+2}{x^3-5x^2+3x+1}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Therefore, }(x-1)\text{ is a common factor of the numerator and denominator.}
\displaystyle \lim \limits_{x \to 1}\frac{x^4-3x^3+2}{x^3-5x^2+3x+1}
\displaystyle =\lim \limits_{x \to 1}\frac{(x-1)(x^3-2x^2-2x-2)}{(x-1)(x^2-4x-1)}
\displaystyle =\lim \limits_{x \to 1}\frac{x^3-2x^2-2x-2}{x^2-4x-1}
\displaystyle =\frac{1^3-2(1)^2-2(1)-2}{1^2-4(1)-1}
\displaystyle =\frac{-5}{-4}
\displaystyle =\frac{5}{4}
\displaystyle \\

\displaystyle \textbf{Question 26: } \lim \limits_{x \to 2}\frac{x^3+3x^2-9x-2}{x^3-x-6}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=2,\text{ the expression }\frac{x^3+3x^2-9x-2}{x^3-x-6}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Therefore, }(x-2)\text{ is a common factor of the numerator and denominator.}
\displaystyle \lim \limits_{x \to 2}\frac{x^3+3x^2-9x-2}{x^3-x-6}
\displaystyle =\lim \limits_{x \to 2}\frac{(x-2)(x^2+5x+1)}{(x-2)(x^2+2x+3)}
\displaystyle =\lim \limits_{x \to 2}\frac{x^2+5x+1}{x^2+2x+3}
\displaystyle =\frac{2^2+5(2)+1}{2^2+2(2)+3}
\displaystyle =\frac{4+10+1}{4+4+3}
\displaystyle =\frac{15}{11}
\displaystyle \\

\displaystyle \textbf{Question 27: } \lim \limits_{x \to 1}\frac{1-x^{-\frac13}}{1-x^{-\frac23}}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=1,\text{ the expression }\frac{1-x^{-\frac13}}{1-x^{-\frac23}}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Factorizing the denominator, we get}
\displaystyle \lim \limits_{x \to 1}\frac{1-x^{-\frac13}}{1-x^{-\frac23}}
\displaystyle =\lim \limits_{x \to 1}\frac{1-x^{-\frac13}}{1^2-\left(x^{-\frac13}\right)^2}
\displaystyle =\lim \limits_{x \to 1}\frac{1-x^{-\frac13}}{(1-x^{-\frac13})(1+x^{-\frac13})}
\displaystyle =\lim \limits_{x \to 1}\frac{1}{1+x^{-\frac13}}
\displaystyle =\frac{1}{1+1}
\displaystyle =\frac{1}{2}
\displaystyle \\

\displaystyle \textbf{Question 28: } \lim \limits_{x \to 3}\frac{x^2-x-6}{x^3-3x^2+x-3}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=3,\text{ the expression }\frac{x^2-x-6}{x^3-3x^2+x-3}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Therefore, }(x-3)\text{ is a common factor of the numerator and denominator.}
\displaystyle \lim \limits_{x \to 3}\frac{x^2-x-6}{x^3-3x^2+x-3}
\displaystyle =\lim \limits_{x \to 3}\frac{x^2-3x+2x-6}{x^2(x-3)+(x-3)}
\displaystyle =\lim \limits_{x \to 3}\frac{x(x-3)+2(x-3)}{(x^2+1)(x-3)}
\displaystyle =\lim \limits_{x \to 3}\frac{(x+2)(x-3)}{(x^2+1)(x-3)}
\displaystyle =\lim \limits_{x \to 3}\frac{x+2}{x^2+1}
\displaystyle =\frac{3+2}{3^2+1}
\displaystyle =\frac{5}{10}
\displaystyle =\frac{1}{2}
\displaystyle \\

\displaystyle \textbf{Question 29: } \lim \limits_{x \to -2}\frac{x^3+x^2+4x+12}{x^3-3x+2}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=-2,\text{ the expression }\frac{x^3+x^2+4x+12}{x^3-3x+2}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Therefore, }(x+2)\text{ is a common factor of the numerator and denominator.}
\displaystyle \lim \limits_{x \to -2}\frac{x^3+x^2+4x+12}{x^3-3x+2}
\displaystyle =\lim \limits_{x \to -2}\frac{(x+2)(x^2-x+6)}{(x+2)(x^2-2x+1)}
\displaystyle =\lim \limits_{x \to -2}\frac{x^2-x+6}{x^2-2x+1}
\displaystyle =\frac{(-2)^2-(-2)+6}{(-2)^2-2(-2)+1}
\displaystyle =\frac{4+2+6}{4+4+1}
\displaystyle =\frac{12}{9}
\displaystyle =\frac{4}{3}
\displaystyle \\

\displaystyle \textbf{Question 30: } \lim \limits_{x \to 1}\frac{x^3+3x^2-6x+2}{x^3+3x^2-3x-1}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=1,\text{ the expression }\frac{x^3+3x^2-6x+2}{x^3+3x^2-3x-1}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Therefore, }(x-1)\text{ is a common factor of the numerator and denominator.}
\displaystyle \lim \limits_{x \to 1}\frac{x^3+3x^2-6x+2}{x^3+3x^2-3x-1}
\displaystyle =\lim \limits_{x \to 1}\frac{(x-1)(x^2+4x-2)}{(x-1)(x^2+4x+1)}
\displaystyle =\lim \limits_{x \to 1}\frac{x^2+4x-2}{x^2+4x+1}
\displaystyle =\frac{1^2+4(1)-2}{1^2+4(1)+1}
\displaystyle =\frac{1+4-2}{1+4+1}
\displaystyle =\frac{3}{6}
\displaystyle =\frac{1}{2}
\displaystyle \\

\displaystyle \textbf{Question 31: } \lim \limits_{x \to 2}\left(\frac{1}{x-2}-\frac{2(2x-3)}{x^3-3x^2+2x}\right).
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=2,\text{ the expression assumes the indeterminate form }\infty-\infty.
\displaystyle \text{Factorizing the second denominator and combining the fractions, we get}
\displaystyle \lim \limits_{x \to 2}\left(\frac{1}{x-2}-\frac{2(2x-3)}{x^3-3x^2+2x}\right)
\displaystyle =\lim \limits_{x \to 2}\left(\frac{1}{x-2}-\frac{2(2x-3)}{x(x^2-3x+2)}\right)
\displaystyle =\lim \limits_{x \to 2}\left(\frac{1}{x-2}-\frac{2(2x-3)}{x(x-1)(x-2)}\right)
\displaystyle =\lim \limits_{x \to 2}\frac{x(x-1)-2(2x-3)}{x(x-1)(x-2)}
\displaystyle =\lim \limits_{x \to 2}\frac{x^2-x-4x+6}{x(x-1)(x-2)}
\displaystyle =\lim \limits_{x \to 2}\frac{x^2-5x+6}{x(x-1)(x-2)}
\displaystyle =\lim \limits_{x \to 2}\frac{(x-2)(x-3)}{x(x-1)(x-2)}
\displaystyle =\lim \limits_{x \to 2}\frac{x-3}{x(x-1)}
\displaystyle =\frac{2-3}{2(2-1)}
\displaystyle =-\frac{1}{2}
\displaystyle \\

\displaystyle \textbf{Question 32: } \lim \limits_{x \to 1}\frac{\sqrt{x^2-1}+\sqrt{x-1}}{\sqrt{x^2-1}},\ x>1.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=1,\text{ the expression }\frac{\sqrt{x^2-1}+\sqrt{x-1}}{\sqrt{x^2-1}}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Since }x^2-1=(x-1)(x+1),\text{ we get}
\displaystyle \lim \limits_{x \to 1}\frac{\sqrt{x^2-1}+\sqrt{x-1}}{\sqrt{x^2-1}}
\displaystyle =\lim \limits_{x \to 1}\left(1+\frac{\sqrt{x-1}}{\sqrt{x^2-1}}\right)
\displaystyle =\lim \limits_{x \to 1}\left(1+\frac{\sqrt{x-1}}{\sqrt{x-1}\sqrt{x+1}}\right)
\displaystyle =\lim \limits_{x \to 1}\left(1+\frac{1}{\sqrt{x+1}}\right)
\displaystyle =1+\frac{1}{\sqrt{1+1}}
\displaystyle =\frac{\sqrt{2}+1}{\sqrt{2}}
\displaystyle \\

\displaystyle \textbf{Question 33: } \lim \limits_{x \to 1}\left(\frac{x-2}{x^2-x}+\frac{1}{x^3-3x^2+2x}\right).
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=1,\text{ the expression assumes the indeterminate form }\infty-\infty.
\displaystyle \text{Factorizing the denominators and combining the fractions, we get}
\displaystyle \lim \limits_{x \to 1}\left(\frac{x-2}{x^2-x}+\frac{1}{x^3-3x^2+2x}\right)
\displaystyle =\lim \limits_{x \to 1}\left(\frac{x-2}{x(x-1)}+\frac{1}{x(x^2-3x+2)}\right)
\displaystyle =\lim \limits_{x \to 1}\left(\frac{x-2}{x(x-1)}+\frac{1}{x(x-1)(x-2)}\right)
\displaystyle =\lim \limits_{x \to 1}\frac{(x-2)^2-1}{x(x-1)(x-2)}
\displaystyle =\lim \limits_{x \to 1}\frac{(x-1)(x-3)}{x(x-1)(x-2)}
\displaystyle =\lim \limits_{x \to 1}\frac{x-3}{x(x-2)}
\displaystyle =\frac{1-3}{1(1-2)}
\displaystyle =2
\displaystyle \\

\displaystyle \textbf{Question 34: } \lim \limits_{x \to 1}\frac{x^7-2x^5+1}{x^3-3x^2+2}.
\displaystyle \textbf{Answer:}
\displaystyle \text{When }x=1,\text{ the expression }\frac{x^7-2x^5+1}{x^3-3x^2+2}\text{ assumes the form }\frac{0}{0}.
\displaystyle \text{Therefore, }(x-1)\text{ is a common factor of the numerator and denominator.}
\displaystyle \lim \limits_{x \to 1}\frac{x^7-2x^5+1}{x^3-3x^2+2}
\displaystyle =\lim \limits_{x \to 1}\frac{(x-1)(x^6+x^5-x^4-x^3-x^2-x-1)}{(x-1)(x^2-2x-2)}
\displaystyle =\lim \limits_{x \to 1}\frac{x^6+x^5-x^4-x^3-x^2-x-1}{x^2-2x-2}
\displaystyle =\frac{1^6+1^5-1^4-1^3-1^2-1-1}{1^2-2(1)-2}
\displaystyle =\frac{-3}{-3}
\displaystyle =1
\displaystyle \\

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