$\lim_{{x \rightarrow -}}\left(\frac{-\pi}{\cos(x)}\right)$ is equal to
$\lim_{{x \rightarrow -}}\left(\frac{-\pi}{\cos(x)}\right)$ is equal to
- $0$
- $\frac{1}{2}$
- $-2$
- $5$
Solution
Let $\ell=\lim _{x \rightarrow 2} \frac{\pi}{2}\left(\frac{2 x-\pi}{\cos x}\right) \quad\left[\right.$ form $\left.\frac{0}{0}\right]$
Use L' Hospital rule $=\lim _{x \rightarrow \pi / 2} \frac{2}{-\sin x}=\frac{2}{-1}=-2$
Asked in: AP EAMCET 2015
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