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  • Using the method of L’ Hospital, evaluate the limit. <img alt="\lim _{x \rightarrow 0} \frac{\log (\sec x)}{1-\sec x}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Clim%20

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

\lim _{x \rightarrow 0} \frac{\log (\sec x)}{1-\sec x}

Option: 1

1


Option: 2

0


Option: 3

-1


Option: 4

Cannot be determined


Answers (1)

best_answer

Apply the L' Hospital's Rule, when   \lim _{x \rightarrow 0} \frac{f(x)}{g(x)}=\frac{0}{0}, \frac{\infty}{\infty}, \text { etc. }   [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 0} \frac{\log (\sec x)}{1-\sec x}=\frac{0}{0}

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

\begin{aligned} & \frac{d}{d x}(\sec x)=\sec x \tan x \\ & \frac{d}{d x}\left(\log _e x\right)=\frac{1}{x} \end{aligned}

Now, derive the following:

\begin{aligned} & \lim _{x \rightarrow 0} \frac{\log (\sec x)}{1-\sec x} \\ & =\lim _{x \rightarrow 0} \frac{\frac{d}{d x} \log (\sec x)}{\frac{d}{d x}(1-\sec x)} \\ & =\lim _{x \rightarrow 0} \frac{\frac{\sec x \tan x}{\sec x}}{-\sec x \tan x} \\ & =-\lim _{x \rightarrow 0} \frac{1}{\sec x} \quad[\tan x \neq 0 \text { as } x \neq 0] \\ & =-\frac{1}{\sec 0} \\ & =-1 \end{aligned}

 

Posted by

Ritika Jonwal

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