Evaluate the following limits: 

\displaystyle \textbf{Question 1: } \lim \limits_{x\to\infty}\frac{(3x-1)(4x-2)}{(x+8)(x-1)}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Here the expression assumes the form }\frac{\infty}{\infty}.\text{ The highest power of }x\text{ in both}
\displaystyle \text{the numerator and denominator is }2.\text{ Therefore, divide each term in both}
\displaystyle \text{the numerator and denominator by }x^2.
\displaystyle \text{When }x\to\infty,\text{ then }\frac{1}{x}\to0.
\displaystyle \therefore \lim \limits_{x\to\infty}\frac{(3x-1)(4x-2)}{(x+8)(x-1)}
\displaystyle =\lim \limits_{x\to\infty}\frac{\left(\frac{3x-1}{x}\right)\left(\frac{4x-2}{x}\right)}{\left(\frac{x+8}{x}\right)\left(\frac{x-1}{x}\right)}
\displaystyle =\lim \limits_{x\to\infty}\frac{\left(3-\frac{1}{x}\right)\left(4-\frac{2}{x}\right)}{\left(1+\frac{8}{x}\right)\left(1-\frac{1}{x}\right)}
\displaystyle =\frac{3\times4}{1\times1}
\displaystyle =12
\displaystyle \\

\displaystyle \textbf{Question 2: } \lim \limits_{x\to\infty}\frac{3x^3-4x^2+6x-1}{2x^3+x^2-5x+7}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Here the expression assumes the form }\frac{\infty}{\infty}.\text{ The highest power of }x\text{ in both}
\displaystyle \text{the numerator and denominator is }3.\text{ Therefore, divide each term in both}
\displaystyle \text{the numerator and denominator by }x^3.
\displaystyle \text{When }x\to\infty,\text{ then }\frac{1}{x},\frac{1}{x^2},\frac{1}{x^3}\to0.
\displaystyle \therefore \lim \limits_{x\to\infty}\frac{3x^3-4x^2+6x-1}{2x^3+x^2-5x+7}
\displaystyle =\lim \limits_{x\to\infty}\frac{3-\frac{4}{x}+\frac{6}{x^2}-\frac{1}{x^3}}{2+\frac{1}{x}-\frac{5}{x^2}+\frac{7}{x^3}}
\displaystyle =\frac{3}{2}
\displaystyle \\

\displaystyle \textbf{Question 3: } \lim \limits_{x\to\infty}\frac{5x^3-6}{\sqrt{9+4x^6}}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Here the expression assumes the form }\frac{\infty}{\infty}.\text{ The highest power of }x\text{ in both}
\displaystyle \text{the numerator and denominator is }3.\text{ Therefore, divide each term in both}
\displaystyle \text{the numerator and denominator by }x^3.
\displaystyle \text{When }x\to\infty,\text{ then }\frac{1}{x^3},\frac{1}{x^6}\to0.
\displaystyle \therefore \lim \limits_{x\to\infty}\frac{5x^3-6}{\sqrt{9+4x^6}}
\displaystyle =\lim \limits_{x\to\infty}\frac{\frac{5x^3-6}{x^3}}{\frac{\sqrt{9+4x^6}}{x^3}}
\displaystyle =\lim \limits_{x\to\infty}\frac{5-\frac{6}{x^3}}{\sqrt{\frac{9}{x^6}+4}}
\displaystyle =\frac{5}{2}
\displaystyle \\

\displaystyle \textbf{Question 4: } \lim \limits_{x\to\infty}\left(\sqrt{x^2+cx}-x\right).
\displaystyle \textbf{Answer:}
\displaystyle \text{This is of the form }\infty-\infty.
\displaystyle \text{Rationalizing the expression,}
\displaystyle \lim \limits_{x\to\infty}\left(\sqrt{x^2+cx}-x\right)=\lim \limits_{x\to\infty}\left(\sqrt{x^2+cx}-x\right)\times\frac{\sqrt{x^2+cx}+x}{\sqrt{x^2+cx}+x}
\displaystyle =\lim \limits_{x\to\infty}\frac{x^2+cx-x^2}{\sqrt{x^2+cx}+x}
\displaystyle =\lim \limits_{x\to\infty}\frac{cx}{\sqrt{x^2+cx}+x}
\displaystyle \text{Here the expression assumes the form }\frac{\infty}{\infty}.\text{ The highest power of }x\text{ in both}
\displaystyle \text{the numerator and denominator is }1.\text{ Therefore, divide each term in both}
\displaystyle \text{the numerator and denominator by }x.
\displaystyle \text{When }x\to\infty,\text{ then }\frac{1}{x}\to0.
\displaystyle =\lim \limits_{x\to\infty}\frac{c}{\sqrt{1+\frac{c}{x}}+1}
\displaystyle =\frac{c}{\sqrt{1}+1}
\displaystyle =\frac{c}{2}
\displaystyle \\

\displaystyle \textbf{Question 5: } \lim \limits_{x\to\infty}\left(\sqrt{x+1}-\sqrt{x}\right).
\displaystyle \textbf{Answer:}
\displaystyle \text{This is of the form }\infty-\infty.
\displaystyle \text{Rationalizing the expression,}
\displaystyle \lim \limits_{x\to\infty}\left(\sqrt{x+1}-\sqrt{x}\right)=\lim \limits_{x\to\infty}\left(\sqrt{x+1}-\sqrt{x}\right)\times\frac{\sqrt{x+1}+\sqrt{x}}{\sqrt{x+1}+\sqrt{x}}
\displaystyle =\lim \limits_{x\to\infty}\frac{x+1-x}{\sqrt{x+1}+\sqrt{x}}
\displaystyle =\lim \limits_{x\to\infty}\frac{1}{\sqrt{x+1}+\sqrt{x}}
\displaystyle =0
\displaystyle \\

\displaystyle \textbf{Question 6: } \lim \limits_{x\to\infty}\left(\sqrt{x^2+7x}-x\right).
\displaystyle \textbf{Answer:}
\displaystyle \text{This is of the form }\infty-\infty.
\displaystyle \text{Rationalizing the expression,}
\displaystyle \lim \limits_{x\to\infty}\left(\sqrt{x^2+7x}-x\right)=\lim \limits_{x\to\infty}\left(\sqrt{x^2+7x}-x\right)\times\frac{\sqrt{x^2+7x}+x}{\sqrt{x^2+7x}+x}
\displaystyle =\lim \limits_{x\to\infty}\frac{x^2+7x-x^2}{\sqrt{x^2+7x}+x}
\displaystyle =\lim \limits_{x\to\infty}\frac{7x}{\sqrt{x^2+7x}+x}
\displaystyle \text{Here the expression assumes the form }\frac{\infty}{\infty}.\text{ The highest power of }x\text{ in both}
\displaystyle \text{the numerator and denominator is }1.\text{ Therefore, divide each term in both}
\displaystyle \text{the numerator and denominator by }x.
\displaystyle \text{When }x\to\infty,\text{ then }\frac{1}{x}\to0.
\displaystyle =\lim \limits_{x\to\infty}\frac{7}{\sqrt{1+\frac{7}{x}}+1}
\displaystyle =\frac{7}{\sqrt{1}+1}
\displaystyle =\frac{7}{2}
\displaystyle \\

\displaystyle \textbf{Question 7: } \lim \limits_{x\to\infty}\frac{x}{\sqrt{4x^2+1}-1}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Here the expression assumes the form }\frac{\infty}{\infty}.\text{ The highest power of }x\text{ in both}
\displaystyle \text{the numerator and denominator is }1.\text{ Therefore, divide each term in both}
\displaystyle \text{the numerator and denominator by }x.
\displaystyle \text{When }x\to\infty,\text{ then }\frac{1}{x},\frac{1}{x^2}\to0.
\displaystyle \therefore \lim \limits_{x\to\infty}\frac{x}{\sqrt{4x^2+1}-1}
\displaystyle =\lim \limits_{x\to\infty}\frac{1}{\sqrt{4+\frac{1}{x^2}}-\frac{1}{x}}
\displaystyle =\frac{1}{\sqrt{4}-0}
\displaystyle =\frac{1}{2}
\displaystyle \\

\displaystyle \textbf{Question 8: } \lim \limits_{n\to\infty}\frac{n^2}{1+2+3+\cdots+n}.
\displaystyle \textbf{Answer:}
\displaystyle \lim \limits_{n\to\infty}\frac{n^2}{1+2+3+\cdots+n}=\lim \limits_{n\to\infty}\frac{n^2}{\frac{n(n+1)}{2}}
\displaystyle \text{Here the expression assumes the form }\frac{\infty}{\infty}.\text{ The highest power of }n\text{ in both}
\displaystyle \text{the numerator and denominator is }2.\text{ Therefore, divide each term in both}
\displaystyle \text{the numerator and denominator by }n^2.
\displaystyle \text{When }n\to\infty,\text{ then }\frac{1}{n}\to0.
\displaystyle =\lim \limits_{n\to\infty}\frac{1}{\frac{1\left(1+\frac{1}{n}\right)}{2}}
\displaystyle =\frac{2}{1}
\displaystyle =2
\displaystyle \\

\displaystyle \textbf{Question 9: } \lim \limits_{x\to\infty}\frac{3x^{-1}+4x^{-2}}{5x^{-1}+6x^{-2}}.
\displaystyle \textbf{Answer:}
\displaystyle \lim \limits_{x\to\infty}\frac{3x^{-1}+4x^{-2}}{5x^{-1}+6x^{-2}}=\lim \limits_{x\to\infty}\frac{\frac{3}{x}+\frac{4}{x^2}}{\frac{5}{x}+\frac{6}{x^2}}
\displaystyle =\lim \limits_{x\to\infty}\frac{3x+4}{5x+6}\qquad\left(\text{Multiplying the numerator and denominator by }x^2\right)
\displaystyle \text{Here the expression assumes the form }\frac{\infty}{\infty}.\text{ The highest power of }x\text{ in both}
\displaystyle \text{the numerator and denominator is }1.\text{ Therefore, divide each term in both}
\displaystyle \text{the numerator and denominator by }x.
\displaystyle \text{When }x\to\infty,\text{ then }\frac{1}{x}\to0.
\displaystyle =\lim \limits_{x\to\infty}\frac{3+\frac{4}{x}}{5+\frac{6}{x}}
\displaystyle =\frac{3}{5}
\displaystyle \\

\displaystyle \textbf{Question 10: } \lim \limits_{x\to\infty}\frac{\sqrt{x^2+a^2}-\sqrt{x^2+b^2}}{\sqrt{x^2+c^2}-\sqrt{x^2+d^2}}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Rationalizing the numerator and denominator,}
\displaystyle \lim \limits_{x\to\infty}\frac{\sqrt{x^2+a^2}-\sqrt{x^2+b^2}}{\sqrt{x^2+c^2}-\sqrt{x^2+d^2}}
\displaystyle =\lim \limits_{x\to\infty}\frac{\left(\sqrt{x^2+a^2}-\sqrt{x^2+b^2}\right)}{\left(\sqrt{x^2+c^2}-\sqrt{x^2+d^2}\right)}\times\frac{\sqrt{x^2+c^2}+\sqrt{x^2+d^2}}{\sqrt{x^2+c^2}+\sqrt{x^2+d^2}}\times\frac{\sqrt{x^2+a^2}+\sqrt{x^2+b^2}}{\sqrt{x^2+a^2}+\sqrt{x^2+b^2}}
\displaystyle =\lim \limits_{x\to\infty}\frac{(x^2+a^2)-(x^2+b^2)}{(x^2+c^2)-(x^2+d^2)}\times\frac{\sqrt{x^2+c^2}+\sqrt{x^2+d^2}}{\sqrt{x^2+a^2}+\sqrt{x^2+b^2}}
\displaystyle =\lim \limits_{x\to\infty}\frac{a^2-b^2}{c^2-d^2}\times\frac{\sqrt{x^2+c^2}+\sqrt{x^2+d^2}}{\sqrt{x^2+a^2}+\sqrt{x^2+b^2}}
\displaystyle \text{Here the expression assumes the form }\frac{\infty}{\infty}.\text{ The highest power of }x\text{ in both}
\displaystyle \text{the numerator and denominator is }1.\text{ Therefore, divide each term in both}
\displaystyle \text{the numerator and denominator by }x.
\displaystyle \text{When }x\to\infty,\text{ then }\frac{1}{x}\to0.
\displaystyle =\lim \limits_{x\to\infty}\frac{a^2-b^2}{c^2-d^2}\times\frac{\sqrt{1+\frac{c^2}{x^2}}+\sqrt{1+\frac{d^2}{x^2}}}{\sqrt{1+\frac{a^2}{x^2}}+\sqrt{1+\frac{b^2}{x^2}}}
\displaystyle =\frac{a^2-b^2}{c^2-d^2}\times\frac{\sqrt{1}+\sqrt{1}}{\sqrt{1}+\sqrt{1}}
\displaystyle =\frac{a^2-b^2}{c^2-d^2}
\displaystyle \\

\displaystyle \textbf{Question 11: } \lim \limits_{n\to\infty}\frac{(n+2)!+(n+1)!}{(n+2)!-(n+1)!}.
\displaystyle \textbf{Answer:}
\displaystyle \lim \limits_{n\to\infty}\frac{(n+2)!+(n+1)!}{(n+2)!-(n+1)!}
\displaystyle =\lim \limits_{n\to\infty}\frac{(n+2)(n+1)!+(n+1)!}{(n+2)(n+1)!-(n+1)!}
\displaystyle =\lim \limits_{n\to\infty}\frac{(n+3)(n+1)!}{(n+1)(n+1)!}
\displaystyle =\lim \limits_{n\to\infty}\frac{n+3}{n+1}
\displaystyle \text{Here the expression assumes the form }\frac{\infty}{\infty}.\text{ The highest power of }n\text{ in both}
\displaystyle \text{the numerator and denominator is }1.\text{ Therefore, divide each term in both}
\displaystyle \text{the numerator and denominator by }n.
\displaystyle \text{When }n\to\infty,\text{ then }\frac{1}{n}\to0.
\displaystyle =\lim \limits_{n\to\infty}\frac{1+\frac{3}{n}}{1+\frac{1}{n}}
\displaystyle =\frac{1}{1}
\displaystyle =1
\displaystyle \\

\displaystyle \textbf{Question 12: } \lim \limits_{x\to\infty}x\left(\sqrt{x^2+1}-\sqrt{x^2-1}\right).
\displaystyle \textbf{Answer:}
\displaystyle \text{Rationalizing the expression,}
\displaystyle =\lim \limits_{x\to\infty}x\left(\sqrt{x^2+1}-\sqrt{x^2-1}\right)\times\frac{\sqrt{x^2+1}+\sqrt{x^2-1}}{\sqrt{x^2+1}+\sqrt{x^2-1}}
\displaystyle =\lim \limits_{x\to\infty}x\left(\frac{(x^2+1)-(x^2-1)}{\sqrt{x^2+1}+\sqrt{x^2-1}}\right)
\displaystyle =\lim \limits_{x\to\infty}\frac{2x}{\sqrt{x^2+1}+\sqrt{x^2-1}}
\displaystyle \text{Here the expression assumes the form }\frac{\infty}{\infty}.\text{ The highest power of }x\text{ in both}
\displaystyle \text{the numerator and denominator is }1.\text{ Therefore, divide each term in both}
\displaystyle \text{the numerator and denominator by }x.
\displaystyle \text{When }x\to\infty,\text{ then }\frac{1}{x},\frac{1}{x^2}\to0.
\displaystyle =\lim \limits_{x\to\infty}\frac{2}{\sqrt{1+\frac{1}{x^2}}+\sqrt{1-\frac{1}{x^2}}}
\displaystyle =\frac{2}{\sqrt{1}+\sqrt{1}}
\displaystyle =1
\displaystyle \\

\displaystyle \textbf{Question 13: } \lim \limits_{x\to\infty}\left(\sqrt{x+1}-\sqrt{x}\right)\sqrt{x+2}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Rationalizing the expression,}
\displaystyle \lim \limits_{x\to\infty}\left(\sqrt{x+1}-\sqrt{x}\right)\sqrt{x+2}
\displaystyle =\lim \limits_{x\to\infty}\left(\sqrt{x+1}-\sqrt{x}\right)\sqrt{x+2}\times\frac{\sqrt{x+1}+\sqrt{x}}{\sqrt{x+1}+\sqrt{x}}
\displaystyle =\lim \limits_{x\to\infty}\frac{\sqrt{x+2}\left((x+1)-x\right)}{\sqrt{x+1}+\sqrt{x}}
\displaystyle =\lim \limits_{x\to\infty}\frac{\sqrt{x+2}}{\sqrt{x+1}+\sqrt{x}}
\displaystyle \text{Here the expression assumes the form }\frac{\infty}{\infty}.\text{ The highest power of }x\text{ in both}
\displaystyle \text{the numerator and denominator is }\frac{1}{2}.\text{ Therefore, divide each term in both}
\displaystyle \text{the numerator and denominator by }\sqrt{x}.
\displaystyle \text{When }x\to\infty,\text{ then }\frac{1}{x}\to0.
\displaystyle =\lim \limits_{x\to\infty}\frac{\sqrt{1+\frac{2}{x}}}{\sqrt{1+\frac{1}{x}}+1}
\displaystyle =\frac{\sqrt{1}}{\sqrt{1}+1}
\displaystyle =\frac{1}{2}
\displaystyle \\

\displaystyle \textbf{Question 14: } \lim \limits_{n\to\infty}\frac{1^2+2^2+\cdots+n^2}{n^3}.
\displaystyle \textbf{Answer:}
\displaystyle \lim \limits_{n\to\infty}\frac{1^2+2^2+\cdots+n^2}{n^3}
\displaystyle =\lim \limits_{n\to\infty}\frac{n(n+1)(2n+1)}{6n^3}
\displaystyle =\lim \limits_{n\to\infty}\frac{(n+1)(2n+1)}{6n^2}
\displaystyle =\lim \limits_{n\to\infty}\frac{\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)}{6}
\displaystyle =\frac{(1+0)(2+0)}{6}
\displaystyle =\frac{2}{6}
\displaystyle =\frac{1}{3}
\displaystyle \\

\displaystyle \textbf{Question 15: } \lim \limits_{n\to\infty}\left(\frac{1}{n^2}+\frac{2}{n^2}+\frac{3}{n^2}+\cdots+\frac{n-1}{n^2}\right).
\displaystyle \textbf{Answer:}
\displaystyle \lim \limits_{n\to\infty}\left(\frac{1}{n^2}+\frac{2}{n^2}+\frac{3}{n^2}+\cdots+\frac{n-1}{n^2}\right)
\displaystyle =\lim \limits_{n\to\infty}\frac{1+2+3+\cdots+(n-1)}{n^2}
\displaystyle =\lim \limits_{n\to\infty}\frac{\frac{n(n-1)}{2}}{n^2}
\displaystyle =\lim \limits_{n\to\infty}\frac{n(n-1)}{2n^2}
\displaystyle =\lim \limits_{n\to\infty}\frac{1-\frac{1}{n}}{2}
\displaystyle =\frac{1}{2}
\displaystyle \\

\displaystyle \textbf{Question 16: } \lim \limits_{n\to\infty}\frac{1^3+2^3+\cdots+n^3}{n^4}.
\displaystyle \textbf{Answer:}
\displaystyle \lim \limits_{n\to\infty}\frac{1^3+2^3+\cdots+n^3}{n^4}
\displaystyle =\lim \limits_{n\to\infty}\frac{\left(\frac{n(n+1)}{2}\right)^2}{n^4}
\displaystyle =\lim \limits_{n\to\infty}\frac{n^2(n+1)^2}{4n^4}
\displaystyle =\lim \limits_{n\to\infty}\frac{(n+1)^2}{4n^2}
\displaystyle =\lim \limits_{n\to\infty}\frac{1}{4}\left(1+\frac{1}{n}\right)^2
\displaystyle =\frac{1}{4}
\displaystyle \\

\displaystyle \textbf{Question 17: } \lim \limits_{n\to\infty}\frac{1^3+2^3+\cdots+n^3}{(n-1)^4}.
\displaystyle \textbf{Answer:}
\displaystyle \lim \limits_{n\to\infty}\frac{1^3+2^3+\cdots+n^3}{(n-1)^4}
\displaystyle =\lim \limits_{n\to\infty}\frac{\left(\frac{n(n+1)}{2}\right)^2}{(n-1)^4}
\displaystyle =\lim \limits_{n\to\infty}\frac{n^2(n+1)^2}{4(n-1)^4}
\displaystyle \text{Here the expression assumes the form }\frac{\infty}{\infty}.\text{ The highest power of }n\text{ in both}
\displaystyle \text{the numerator and denominator is }4.\text{ Therefore, divide each term in both}
\displaystyle \text{the numerator and denominator by }n^4.
\displaystyle \text{When }n\to\infty,\text{ then }\frac{1}{n}\to0.
\displaystyle =\lim \limits_{n\to\infty}\frac{1\left(1+\frac{1}{n}\right)^2}{4\left(1-\frac{1}{n}\right)^4}
\displaystyle =\frac{1(1+0)^2}{4(1-0)^4}
\displaystyle =\frac{1}{4}
\displaystyle \\

\displaystyle \textbf{Question 18: } \lim \limits_{x\to\infty}\sqrt{x}\left(\sqrt{x+1}-\sqrt{x}\right).
\displaystyle \textbf{Answer:}
\displaystyle \lim \limits_{x\to\infty}\sqrt{x}\left(\sqrt{x+1}-\sqrt{x}\right)
\displaystyle =\lim \limits_{x\to\infty}\sqrt{x}\left(\sqrt{x+1}-\sqrt{x}\right)\times\frac{\sqrt{x+1}+\sqrt{x}}{\sqrt{x+1}+\sqrt{x}}
\displaystyle =\lim \limits_{x\to\infty}\frac{\sqrt{x}\left((x+1)-x\right)}{\sqrt{x+1}+\sqrt{x}}
\displaystyle =\lim \limits_{x\to\infty}\frac{\sqrt{x}}{\sqrt{x+1}+\sqrt{x}}
\displaystyle \text{Here the expression assumes the form }\frac{\infty}{\infty}.\text{ The highest power of }x\text{ in both}
\displaystyle \text{the numerator and denominator is }\frac{1}{2}.\text{ Therefore, divide each term in both}
\displaystyle \text{the numerator and denominator by }\sqrt{x}.
\displaystyle \text{When }x\to\infty,\text{ then }\frac{1}{x}\to0.
\displaystyle =\lim \limits_{x\to\infty}\frac{1}{\sqrt{1+\frac{1}{x}}+1}
\displaystyle =\frac{1}{2}
\displaystyle \\

\displaystyle \textbf{Question 19: } \lim \limits_{n\to\infty}\left(\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\cdots+\frac{1}{3^n}\right).
\displaystyle \textbf{Answer:}
\displaystyle \lim \limits_{n\to\infty}\left(\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\cdots+\frac{1}{3^n}\right)
\displaystyle =\lim \limits_{n\to\infty}\left[\frac{1}{3}\left(1+\frac{1}{3}+\frac{1}{3^2}+\cdots+\frac{1}{3^{\,n-1}}\right)\right]
\displaystyle =\lim \limits_{n\to\infty}\left[\frac{\frac{1}{3}\left(1-\frac{1}{3^n}\right)}{1-\frac{1}{3}}\right]
\displaystyle =\lim \limits_{n\to\infty}\frac{1}{2}\left(1-\frac{1}{3^n}\right)
\displaystyle =\frac{1}{2}(1-0)
\displaystyle =\frac{1}{2}
\displaystyle \\

\displaystyle \textbf{Question 20: } \lim \limits_{x\to\infty}\frac{x^4+7x^3+46x+a}{x^4+6},\text{ where }a\text{ is a non-zero real number}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Here the expression assumes the form }\frac{\infty}{\infty}.\text{ The highest power of }x\text{ in both}
\displaystyle \text{the numerator and denominator is }4.\text{ Therefore, divide each term in both}
\displaystyle \text{the numerator and denominator by }x^4.
\displaystyle \text{When }x\to\infty,\text{ then }\frac{1}{x},\frac{1}{x^2},\frac{1}{x^3},\frac{1}{x^4}\to0.
\displaystyle =\lim \limits_{x\to\infty}\frac{1+\frac{7}{x}+\frac{46}{x^3}+\frac{a}{x^4}}{1+\frac{6}{x^4}}
\displaystyle =\frac{1+0+0+0}{1+0}
\displaystyle =1
\displaystyle \\

\displaystyle \textbf{Question 21: } f(x)=\frac{ax^2+b}{x^2+1},\ \lim \limits_{x\to0}f(x)=1\text{ and}
\displaystyle \lim \limits_{x\to\infty}f(x)=1,\text{ then prove that }f(-2)=f(2)=1.
\displaystyle \textbf{Answer:}
\displaystyle f(x)=\frac{ax^2+b}{x^2+1}
\displaystyle \text{Given }\lim \limits_{x\to0}f(x)=1.
\displaystyle \Rightarrow \lim \limits_{x\to0}\frac{ax^2+b}{x^2+1}=1
\displaystyle \Rightarrow \frac{a(0)^2+b}{(0)^2+1}=1
\displaystyle \Rightarrow b=1
\displaystyle \text{Also, }\lim \limits_{x\to\infty}f(x)=1.
\displaystyle \Rightarrow \lim \limits_{x\to\infty}\frac{ax^2+b}{x^2+1}=1
\displaystyle \text{Here the expression assumes the form }\frac{\infty}{\infty}.\text{ The highest power of }x\text{ in both}
\displaystyle \text{the numerator and denominator is }2.\text{ Therefore, divide each term in both}
\displaystyle \text{the numerator and denominator by }x^2.
\displaystyle \text{When }x\to\infty,\text{ then }\frac{1}{x^2}\to0.
\displaystyle \Rightarrow \lim \limits_{x\to\infty}\frac{a+\frac{b}{x^2}}{1+\frac{1}{x^2}}=1
\displaystyle \Rightarrow \frac{a+0}{1+0}=1
\displaystyle \Rightarrow a=1
\displaystyle \therefore f(x)=\frac{x^2+1}{x^2+1}=1
\displaystyle \text{Thus, }f(x)\text{ is a constant function.}
\displaystyle \therefore f(-2)=f(2)=1
\displaystyle \\

\displaystyle \textbf{Question 22: } \text{Show that }\lim \limits_{x\to\infty}\left(\sqrt{x^2+x+1}-x\right)\neq
\displaystyle \lim \limits_{x\to\infty}\left(\sqrt{x^2+1}-x\right).
\displaystyle \textbf{Answer:}
\displaystyle \text{LHS:}
\displaystyle \lim \limits_{x\to\infty}\left(\sqrt{x^2+x+1}-x\right)
\displaystyle =\lim \limits_{x\to\infty}\left(\sqrt{x^2+x+1}-x\right)\times\frac{\sqrt{x^2+x+1}+x}{\sqrt{x^2+x+1}+x}
\displaystyle =\lim \limits_{x\to\infty}\frac{x^2+x+1-x^2}{\sqrt{x^2+x+1}+x}
\displaystyle =\lim \limits_{x\to\infty}\frac{x+1}{\sqrt{x^2+x+1}+x}
\displaystyle \text{Here the expression assumes the form }\frac{\infty}{\infty}.\text{ The highest power of }x\text{ in both}
\displaystyle \text{the numerator and denominator is }1.\text{ Therefore, divide each term in both}
\displaystyle \text{the numerator and denominator by }x.
\displaystyle \text{When }x\to\infty,\text{ then }\frac{1}{x},\frac{1}{x^2}\to0.
\displaystyle =\lim \limits_{x\to\infty}\frac{1+\frac{1}{x}}{\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}+1}
\displaystyle =\frac{1}{\sqrt{1}+1}
\displaystyle =\frac{1}{2}
\displaystyle \text{RHS:}
\displaystyle \lim \limits_{x\to\infty}\left(\sqrt{x^2+1}-x\right)
\displaystyle =\lim \limits_{x\to\infty}\left(\sqrt{x^2+1}-x\right)\times\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x}
\displaystyle =\lim \limits_{x\to\infty}\frac{x^2+1-x^2}{\sqrt{x^2+1}+x}
\displaystyle =\lim \limits_{x\to\infty}\frac{1}{\sqrt{x^2+1}+x}
\displaystyle =\lim \limits_{x\to\infty}\frac{\frac{1}{x}}{\sqrt{1+\frac{1}{x^2}}+1}
\displaystyle =0
\displaystyle \text{Therefore, }\lim \limits_{x\to\infty}\left(\sqrt{x^2+x+1}-x\right)\neq\lim \limits_{x\to\infty}\left(\sqrt{x^2+1}-x\right).
\displaystyle \\

\displaystyle \textbf{Question 23: } \lim \limits_{x\to-\infty}\left(\sqrt{4x^2-7x}+2x\right).
\displaystyle \textbf{Answer:}
\displaystyle \text{Given }\lim \limits_{x\to-\infty}\left(\sqrt{4x^2-7x}+2x\right).
\displaystyle \text{Let }x=-m.\text{ When }x\to-\infty,\text{ then }m\to\infty.
\displaystyle \therefore \lim \limits_{m\to\infty}\left(\sqrt{4m^2+7m}-2m\right)
\displaystyle =\lim \limits_{m\to\infty}\left(\sqrt{4m^2+7m}-2m\right)\times\frac{\sqrt{4m^2+7m}+2m}{\sqrt{4m^2+7m}+2m}
\displaystyle =\lim \limits_{m\to\infty}\frac{4m^2+7m-4m^2}{\sqrt{4m^2+7m}+2m}
\displaystyle =\lim \limits_{m\to\infty}\frac{7m}{\sqrt{4m^2+7m}+2m}
\displaystyle \text{Here the expression assumes the form }\frac{\infty}{\infty}.\text{ The highest power of }m\text{ in both}
\displaystyle \text{the numerator and denominator is }1.\text{ Therefore, divide each term in both}
\displaystyle \text{the numerator and denominator by }m.
\displaystyle \text{When }m\to\infty,\text{ then }\frac{1}{m}\to0.
\displaystyle =\lim \limits_{m\to\infty}\frac{7}{\sqrt{4+\frac{7}{m}}+2}
\displaystyle =\frac{7}{\sqrt{4}+2}
\displaystyle =\frac{7}{4}
\displaystyle \\

\displaystyle \textbf{Question 24: } \lim \limits_{x\to-\infty}\left(\sqrt{x^2-8x}+x\right).
\displaystyle \textbf{Answer:}
\displaystyle \text{Given }\lim \limits_{x\to-\infty}\left(\sqrt{x^2-8x}+x\right).
\displaystyle \text{Let }x=-m.\text{ When }x\to-\infty,\text{ then }m\to\infty.
\displaystyle \therefore \lim \limits_{m\to\infty}\left(\sqrt{m^2+8m}-m\right)
\displaystyle =\lim \limits_{m\to\infty}\left(\sqrt{m^2+8m}-m\right)\times\frac{\sqrt{m^2+8m}+m}{\sqrt{m^2+8m}+m}
\displaystyle =\lim \limits_{m\to\infty}\frac{m^2+8m-m^2}{\sqrt{m^2+8m}+m}
\displaystyle =\lim \limits_{m\to\infty}\frac{8m}{\sqrt{m^2+8m}+m}
\displaystyle \text{Here the expression assumes the form }\frac{\infty}{\infty}.\text{ The highest power of }m\text{ in both}
\displaystyle \text{the numerator and denominator is }1.\text{ Therefore, divide each term in both}
\displaystyle \text{the numerator and denominator by }m.
\displaystyle \text{When }m\to\infty,\text{ then }\frac{1}{m}\to0.
\displaystyle =\lim \limits_{m\to\infty}\frac{8}{\sqrt{1+\frac{8}{m}}+1}
\displaystyle =\frac{8}{\sqrt{1}+1}
\displaystyle =4
\displaystyle \\

\displaystyle \textbf{Question 25: } \text{Evaluate } \lim \limits_{n\to\infty}\frac{1^4+2^4+\cdots+n^4}{n^5}-\lim \limits_{n\to\infty}\frac{1^3+2^3+\cdots+n^3}{n^5}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Using }\sum \limits_{k=1}^{n}k^4=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}\text{ and }\sum \limits_{k=1}^{n}k^3=\frac{n^2(n+1)^2}{4},
\displaystyle \lim \limits_{n\to\infty}\frac{1^4+2^4+\cdots+n^4}{n^5}-\lim \limits_{n\to\infty}\frac{1^3+2^3+\cdots+n^3}{n^5}
\displaystyle =\lim \limits_{n\to\infty}\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30n^5}-\lim \limits_{n\to\infty}\frac{n^2(n+1)^2}{4n^5}
\displaystyle =\frac{1}{30}\lim \limits_{n\to\infty}\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)\left(3+\frac{3}{n}-\frac{1}{n^2}\right)-\frac{1}{4}\lim \limits_{n\to\infty}\frac{1}{n}\left(1+\frac{1}{n}\right)^2
\displaystyle =\frac{1}{30}(1+0)(2+0)(3+0-0)-\frac{1}{4}(0)(1+0)^2
\displaystyle =\frac{1}{5}
\displaystyle \\

\displaystyle \textbf{Question 26: } \text{Evaluate }\lim \limits_{n\to\infty}\frac{1\cdot2+2\cdot3+3\cdot4+\cdots+n(n+1)}{n^3}.
\displaystyle \textbf{Answer:}
\displaystyle \lim \limits_{n\to\infty}\frac{1\cdot2+2\cdot3+3\cdot4+\cdots+n(n+1)}{n^3}
\displaystyle =\lim \limits_{n\to\infty}\frac{\sum \limits_{k=1}^{n}k(k+1)}{n^3}
\displaystyle =\lim \limits_{n\to\infty}\frac{\sum \limits_{k=1}^{n}k^2+\sum \limits_{k=1}^{n}k}{n^3}
\displaystyle =\lim \limits_{n\to\infty}\frac{\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}}{n^3}
\displaystyle =\lim \limits_{n\to\infty}\frac{n(n+1)\left((2n+1)+3\right)}{6n^3}
\displaystyle =\lim \limits_{n\to\infty}\frac{n(n+1)(n+2)}{3n^3}
\displaystyle =\frac{1}{3}\lim \limits_{n\to\infty}\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)
\displaystyle =\frac{1}{3}(1+0)(1+0)
\displaystyle =\frac{1}{3}
\displaystyle \\

 

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