limits

Question

Find the value of: $$\lim_{\theta \to \pi/2} \left ( 1 - 5 \cot\theta \right )^{\tan\theta}$$

• A. $e^{5}$
   
• B. $e^{-5}$
   
• C. $e^{1/5}$
   
• D. $e^{-1/5}$

Comments

is it b) ?

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i dont know the answer...it was yesterday's qstn in CBT

Answer 1

Since it is 0/0 form L'Hôpital's rule can be used. $\require{cancel}$

Let $\displaystyle y = \lim_{\theta \to \pi/2} (1 - 5 \cot\theta)^{\tan\theta}$

Take $\log$.

$$\begin{align*}
\log y &= \log \Biggl ( \lim_{\theta \to \pi/2} (1 - 5 \cot\theta)^{\tan\theta} \Biggr ) \\[1em]
&= \lim_{\theta \to \pi/2} \Biggl ( \log (1 - 5 \cot\theta)^{\tan\theta} \Biggr ) & \left \{ \substack{\log \text{ of } \lim \,=\, \lim \text{ of } \log\\\text{provided the limit exists.}} \right. \\[1em]
&= \lim_{\theta \to \pi/2} \tan\theta \cdot \log(1 - 5 \cot\theta) & \left\{\log\left(x^y\right) = y\cdot\log x\right.
\end{align*}$$

Now, since $\tan(\pi/2) = \infty$ and $\log(1-5\cot\pi/2) = 0$, we have a $\infty \cdot 0$ form.  We need to convert this to a $0/0$ form to apply the L'Hôpital's rule.

$$\begin{align*}
\log y &= \lim_{\theta \to \pi/2} \frac{\log(1-5 \cot\theta)}{\cot \theta} \\[1em]
&= \lim_{\theta \to \pi/2} \frac{-5 \cdot \cancel{(-\csc^2\theta)}}{(1-5\cot\theta)\cancel{(-\csc^2\theta)}} & \left \{\text{L'Hôpital's rule} \right. \\[1em]
&= \lim_{\theta \to \pi/2} \frac{-5}{1-5 \cot\theta} \\[1em]
\log y &= -5 \\[1em]
\implies y &= e^{-5}
\end{align*}$$

Answer 2

whenever u have to calculate a limit like Limit_x-->a f(x)= 1 and limit _x-->a g(x)=infinity , and if u have to calculate limit _x-->af(x) ^g(x) , so it actually reduces to 1 ^infinity form

then u can apply this  e ^{limit _x-->a g(x) (f(x)-1)}

So here likewise will see g(x)=tanx and f(x)= 1-5cotx