_ 0 e^ +e^ - -2 ^2 is equal to - Vidyadip

MCQ

$\lim _{\theta \rightarrow 0} \frac{e^\theta+e^{-\theta}-2}{\sin ^2 \theta}$ is equal to

✓
1
B
$0$
C
-1
D
$\sqrt{-2}$

✓

Answer

Correct option: A.

1

(A)
$\lim _{\theta \rightarrow 0} \frac{e^\theta+e^{-\theta}-2}{\sin ^2 \theta}=\lim _{\theta \rightarrow 0} \frac{e^\theta+\frac{1}{e^\theta}-2}{\sin ^2 \theta}$
$=\lim _{\theta \rightarrow 0} \frac{\left[e^\theta-1\right]^2}{e^\theta \sin ^2 \theta}$
$=\lim _{\theta \rightarrow 0} \frac{\frac{\left(e^\theta-1\right)^2}{\theta^2}}{e^\theta \cdot \frac{\sin ^2 \theta}{\theta^2}}$
$=\frac{[\log e ]^2}{1}-1$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "MCQ" from Limits→Limits — all question types→Maths STD 11 Science question papers→Maharashtra Board STD 11 Science question papers→Maharashtra Board question paper generator→

Similar questions

If $\left(\frac{3}{2}, \frac{5}{2}\right)$ is the midpoint of line segment intercepted by a line between axes, the equation of the line is

Three ships A, B and C sail from England to India. If the ratio of their arriving safely are 2 : 5, 3 : 7 and 6 : 11 respectively, then the probability of all the ships for arriving safely is

If the functions $f, g, h$ are defined from the sets of real numbers $R$ to $R$ such that
$f(x)=x^2-1, g(x)=\sqrt{x^2+1}, h(x)=\left\{\begin{array}{l}0, \text { if }{ }_{x=0} \\x, \text { if }{ }_{x>0} ,\end{array}\right.$
then the composite function $(hofog)(x)=$

In how many ways can a committee of $5$ be made out of $6$ men and $4$ women containing at least one women?

If $\tan\Big(\frac\pi4+\text{x}\Big)+\tan\Big(\frac\pi4-\text{x}\Big)=\text{a},$ then $\tan^2\Big(\frac\pi4+\text{x}\Big)+\tan^2\Big(\frac\pi4-\text{x}\Big)=$

If $f (x)=\left\{\begin{array}{r}\frac{x-|x|}{x} ; \text { when } x \neq 0 \\ 2 ; \text { when } x=0\end{array}\right.$, then

$\frac{\cos 8^{\circ}-\sin 8^{\circ}}{\cos 8^{\circ}+\sin 8^{\circ}}$ is equal to

$\sec50^{\circ}+\tan50^{\circ}$is equal to

There are 6 positive and 8 negative numbers. From these four numbers are chosen at random and multiplied. Then the probability, that the product is negative number, is

$\tan \frac{2 \pi}{5}-\tan \frac{\pi}{15}-\sqrt{3} \tan \frac{2 \pi}{5} \tan \frac{\pi}{15}$ is equal to