mathop {lim }limits_{x to pi /2} frac{{{a^{co | vedclass

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

Question

(A) Let $L = \mathop {\lim }\limits_{x 	o \pi /2} \frac{{{a^{\cot x}} - {a^{\cos x}}}}{{\cot x - \cos x}}$.We can factor out ${a^{\cos x}}$ from the

Vedclass · Accepted Answer

$\log a$

Vedclass · Answer

(A) Let $L = \mathop {\lim }\limits_{x 	o \pi /2} \frac{{{a^{\cot x}} - {a^{\cos x}}}}{{\cot x - \cos x}}$.We can factor out ${a^{\cos x}}$ from the numerator:$L = \mathop {\lim }\limits_{x 	o \pi /2} \frac{{{a^{\cos x}}({a^{\cot x - \cos x}} - 1)}}{{\cot x - \cos x}}$.As $x 	o \pi /2$,$\cos x 	o 0$,so ${a^{\cos x}} 	o {a^0} = 1$.Using the standard limit $\mathop {\lim }\limits_{u 	o 0} \frac{{{a^u} - 1}}{u} = \log a$,where $u = \cot x - \cos x$ (note that as $x 	o \pi /2$,$u 	o 0 - 0 = 0$):$L = 1 	imes \log a = \log a$.

$\mathop {\lim }\limits_{x \to \pi /2} \frac{{{a^{\cot x}} - {a^{\cos x}}}}{{\cot x - \cos x}} = $

Similar Questions

$\mathop {\lim }\limits_{n \to \infty } \cos \left( {\pi \sqrt {{n^2} + n} } \right)$,where $n \in I$,is

If $\lim_{x \to 0} \frac{(4^x - 1)^3}{\tan(\frac{x}{4}) \log(1 + \frac{x^2}{3})} = 96(\log a)^b$, then $(a + b) = $

If $\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{a}{x} - \frac{4}{{{x^2}}}} \right)^{2x}} = {e^3},$ then $a$ is equal to

If $\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{{a^{1/x}} + b}}{c}} \right)^x} = d$ (where $d$ is a non-zero finite value),then $(b + 1) \log_a d$ is equal to:

$\mathop {Lim}\limits_{x \to \infty } \frac{2 + 2x + \sin 2x}{(2x + \sin 2x)e^{\sin x}}$ is :

Vedclass Test Series

Exam Paper Generator

Online Exam Module