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  • Using the method of L’ Hospital, evaluate the limit. <img alt="\lim _{x \rightarrow 1} \frac{\log x^{\sin z \cos z}}{x-1}" src="https://entrancecorner.oncodecogs

Using the method of L’ Hospital, evaluate the limit.

\lim _{x \rightarrow 1} \frac{\log x^{\sin z \cos z}}{x-1}

Option: 1

\frac{\sin ^2 z}{2}


Option: 2

\sin z


Option: 3

\frac{\sin 2 z}{2}


Option: 4

Cannot be determined


Answers (1)

Apply the L' Hospital's Rule  \lim _{x \rightarrow 0} \frac{f(x)}{g(x)}=\frac{0}{0}, \frac{\infty}{\infty}, \text { etc. } , when [an intermediate form] in the following way

  • Differentiate \lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\lim _{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)}, provided that the limit \lim _{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)} exists.

  • Otherwise, go on differentiating till the limit with the determinate form is achieved.

The provided limit is

\lim _{x \rightarrow 1} \frac{\log x^{\sin z \cos z}}{x-1}=\frac{0}{0}

Here use the L’ Hospital rule and also refer to the following formula.

\frac{d}{d x}\left(\log _e x\right)=\frac{1}{x}

Now, derive the following:

\begin{aligned} & \lim _{x \rightarrow 1} \frac{\log x^{\sin z \cos z}}{x-1} \\ & =\lim _{x \rightarrow 1} \frac{(\sin z \cos z) \log x}{x-1} \\ & =(\sin z \cos z) \lim _{x \rightarrow 1} \frac{\frac{d}{d x}(\log x)}{\frac{d}{d x}(x-1)} \\ & =(\sin z \cos z) \lim _{x \rightarrow 1} \frac{x}{1} \\ & =(\sin z \cos z) \times 1 \\ & =\frac{1}{2} \times 2 \sin z \cos z \\ \end{aligned}

\frac{\sin 2 z}{2}

Posted by

Sumit Saini

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