Let mathop {Lim}limits_{x to 0} sec^{-1} left | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) We know that $\mathop {Lim}\limits_{x 	o 0} \frac{\sin x}{x} = 1$,so $\mathop {Lim}\limits_{x 	o 0} \frac{x}{\sin x} = 1$. Thus,$l = \sec^{-1}(1

Vedclass · Accepted Answer

$l$ and $m$ both exist

Vedclass · Answer

(C) We know that $\mathop {Lim}\limits_{x 	o 0} \frac{\sin x}{x} = 1$,so $\mathop {Lim}\limits_{x 	o 0} \frac{x}{\sin x} = 1$. Thus,$l = \sec^{-1}(1) = 0$. Similarly,$\mathop {Lim}\limits_{x 	o 0} \frac{	an x}{x} = 1$,so $\mathop {Lim}\limits_{x 	o 0} \frac{x}{	an x} = 1$. Thus,$m = \sec^{-1}(1) = 0$. Since both limits exist and are equal to $0$,$l$ and $m$ both exist.

Let $\mathop {Lim}\limits_{x \to 0} \sec^{-1} \left( \frac{x}{\sin x} \right) = l$ and $\mathop {Lim}\limits_{x \to 0} \sec^{-1} \left( \frac{x}{\tan x} \right) = m$,then

Similar Questions

The principal value of $\sin^{-1} \left( \sin \frac{5\pi}{3} \right)$ is

Find the principal values of $\sin ^{-1}\left(-\frac{1}{2}\right)$.

If $\cos ^{-1}\left(\frac{1}{2}\right)=\cot \left(\cos ^{-1} x\right)$, then the value of $x$ is

Write $\cot ^{-1}\left(\frac{1}{\sqrt{x^{2}-1}}\right), x>1$ in the simplest form.

If $\sin (\cot ^{ - 1}(x + 1)) = \cos (\tan ^{ - 1}x)$,then $x =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module