Solution.

Let \(f(x) = x^3 - 5x^2 + 6x - 2\hspace {0.3cm}\) and \(\hspace {0.4cm} g(x) = x^5 - 3x^4 - 7x^2 + 9\). Then \(\lim \limits _{x\rightarrow 1}f(x) = f(1) = 0\). Also \(\lim \limits _{x\rightarrow 1}g(x) = g(1) = 0\)

\begin {align*} \implies \hspace {0.5cm} \lim \limits _{x\rightarrow 1}\dfrac {f(x)}{g(x)}= & \lim \limits _{x\rightarrow 1}\dfrac {f'(x)}{g'(x)} = \dfrac {f'(1)}{g'(1)}\\ & = \dfrac {-1}{2}=\dfrac {1}{2}\\ \end {align*}

If \(f\) and \(g\) are two functions having the property that \(\lim \limits _{x\rightarrow c}f(x) = \pm \infty \) and \(\lim \limits _{x\rightarrow c}g(x) = \pm \infty \), then the function \(\dfrac {f}{g}\) is said to be indeterminate form \(\dfrac {\infty }{\infty }\). Thus, \(\lim \limits _{x\rightarrow c}\dfrac {f(x)}{g(x)} = \lim \limits _{x\rightarrow c}\dfrac {f'(x)}{g'(x)}\).

Solution.

\(\lim \limits _{x\rightarrow +\infty } e^x = + \infty \) and the \(\lim \limits _{x\rightarrow +\infty }x = +\infty \).

\(\implies \hspace {0.3cm} \lim \limits _{x\rightarrow +\infty }\dfrac {f(x)}{g(x)}\) is of indeterminate form \(\Big (\dfrac {\infty }{\infty }\Big )\). Thus \[\lim \limits _{x\rightarrow +\infty }\dfrac {f(x)}{g(x)} = \lim \limits _{x\rightarrow +\infty }\dfrac {f'(x)}{g'(x)} = \lim \limits _{x\rightarrow +\infty }\frac {e^x}{1} = +\infty \]

Solution.

\(\lim \limits _{x\rightarrow 0} \dfrac {\cot x}{\cot 2x}\hspace {1cm} \Bigg (\dfrac {\infty }{\infty }\Bigg )\)

\begin {align*} \lim \limits _{x\rightarrow 0} \dfrac {\cot x}{\cot 2x} & = \lim \limits _{x\rightarrow 0}\dfrac {-\csc ^2x}{-2\csc ^22x}\hspace {1cm} \dfrac {\infty }{\infty } \end {align*}

\[\lim \limits _{x\rightarrow c}\dfrac {f(x)}{g(x)} = \lim \limits _{x\rightarrow c}\dfrac {f'(x)}{g'(x)}= \lim \limits _{x\rightarrow c}\dfrac {f''(x)}{g''(x)}=\lim \limits _{x\rightarrow 0}\dfrac {2\cot x \hspace {0.1cm}\csc x}{8\cot 2x\hspace {0.1cm}\csc 2x}\]

Note that at every stage of trying to find the limit we are getting the indeterminate form \(\Big (\dfrac {\infty }{\infty }\Big )\). Thus, we may need to rewrite the limit as \begin {align*} \lim \limits _{x\rightarrow 0}\dfrac {\cot x}{\cot 2x} & = \lim \limits _{x\rightarrow 0}\dfrac {\dfrac {1}{\tan x}}{\dfrac {1}{\tan 2x}}\\\\ & = \lim \limits _{x\rightarrow 0}\dfrac {\tan 2x}{\tan x}\hspace {1cm} \Big (\dfrac {0}{0}\Big )\\\\ & = \lim \limits _{x\rightarrow 0}\dfrac {2\sec ^22x}{\sec ^2x}\\ & = \frac {2}{1}\\ & = 2\\ \end {align*}

Solution. \(\hspace {0.3cm}\lim \limits _{x\rightarrow \infty }x^2\cdot e^{-x} \hspace {1cm} (\infty \cdot 0)\) \begin {align*} \lim \limits _{x\rightarrow \infty }x^2\cdot e^{-x} & = \lim \limits _{x\rightarrow \infty } \dfrac {x^2}{e^x}\hspace {1cm} \Big (\dfrac {\infty }{\infty }\Big )\\\\ & = \lim \limits _{x\rightarrow \infty }\dfrac {2x}{e^x}\hspace {1cm} \Big (\dfrac {\infty }{\infty }\Big )\\\\ & = \lim \limits _{x\rightarrow \infty } \dfrac {2}{e^x} = 0\\ \end {align*}

Solution. Let \(\hspace {0.5cm}y = \lim \limits _{x\rightarrow 0} \Bigg (1 + \dfrac {1}{x}\Bigg )^x\) Then

\begin {align*} \ln y & = \ln \Bigg [\lim \limits _{x\rightarrow 0} \Big (1 + \dfrac {1}{x}\Big )^x\Bigg ] = \lim \limits _{x\rightarrow 0}\Bigg [\ln \Big (1 + \dfrac {1}{x}\Big )^x\Bigg ]\\\\ & = \lim \limits _{x\rightarrow 0}\Bigg [ x \ln \Big (1 + \dfrac {1}{x}\Big )^x\Bigg ]\hspace {2cm} (0\cdot \infty )\\\\ & = \lim \limits _{x\rightarrow 0}\hspace {0.1cm} \dfrac {\ln \Big ( 1 + \dfrac {1}{x}\Big )}{\dfrac {1}{x}}\hspace {2cm} \dfrac {\infty }{\infty } \end {align*}

\begin {align*} \ln y & = \lim \limits _{x\rightarrow 0}\hspace {0.1cm} \dfrac {\dfrac {1}{1 + 1/x} \Big (-\dfrac {1}{x^2}\Big )}{\Big (-\dfrac {1}{x^2}\Big )}= \lim \limits _{x\rightarrow 0} \hspace {0.1cm}\dfrac {1}{1 + \dfrac {1}{x}}\\ & = 0 \end {align*}

\[\implies \hspace {0.5cm} \ln y = 0\hspace {0.4cm} \implies \hspace {0.4cm} y = e^0 = 1\]

\(\therefore \hspace {0.5cm} \lim \limits _{x\rightarrow 0} \Big (1 + \dfrac {1}{x}\Big )^x = 1\)