Question

$\underset{x \rightarrow \frac{\pi}{2}^{-}}{Lim}$ [1 + (cos x)^{cos x}]² is equal to

• A. Does not exist
• B. 1
• C. e
• D. 4

Answer 1

Answer: (D)

4

Answer 2

$\underset{x \rightarrow \frac{\pi}{2}^{-}}{Lim}$ [1 + (cos x)^cosx]²

y = $\underset{x \rightarrow \frac{\pi}{2}^{-}}{Lim}$ (cos x)^{cos x}

log(y) = $\underset{x \rightarrow \frac{\pi}{2}^{-}}{Lim}$ (cos x) log cos x

log (y) = $\frac{\log(\cos x)}{\sec(x)}$ (∞/∞) L'hospital

log(y) = $\underset{x \rightarrow \frac{\pi}{2}^{-}}{Lim}$ $\frac{1}{\cos x}$× – $\frac{\sin x}{\sec x\tan x}$

= $\underset{x \rightarrow \frac{\pi}{2}^{-}}{Lim}$ – cos x = 0

y = e⁰ = 1

Now limit is (1 + 1)² = 2² = 4