The value of undersetx arrow 0+ textLim (e( sin x)x-1-ex ln ( sin x)/x ln ( sin x)) is equal to

Question

The value of undersetx arrow 0+ textLim (e( sin x)x-1-ex ln ( sin x)/x ln ( sin x)) is equal to (A) 0 (B) 1 (C) e (D) e -1

Tardigrade Admin · Accepted Answer

Let ( sin x)x=t text as x arrow 0, t arrow 1 ∴ underset t arrow 1 textLim ( e t -1- e ln t / ln t ) ∴ e ln t operatornameLim t arrow 1 ( e t -1- ln t -1/ ln t )= underset t arrow 1 textLim ( t -1- ln t / ln t )= underset t arrow 1 textLim(1-(1/ t )/(1) t )=0 .

Q. The value of $x \to 0^{+} Lim \frac{e ^{(s i n x)^{x} - 1} - e ^{x l n (s i n x)}}{x l n ( s i n x )}$ is equal to

0

1

$e$

$e - 1$

Let $(sin x)^{x} = t$
$as x \to 0, t \to 1$
$∴ t \to 1 Lim \frac{e ^{t - 1} - e ^{l n t}}{l n t}$
$∴ e^{l n t} Lim_{t \to 1} \frac{e ^{t - 1 - l n t} - 1}{l n t} = t \to 1 Lim \frac{t - 1 - l n t}{l n t} = t \to 1 Lim \frac{1 - \frac{1}{t}}{\frac{1}{t}} = 0.$

Q. The value of x→0+Lim​xln(sinx)e(sinx)x−1−exln(sinx)​ is equal to

0

1

e

e−1

Let (sinx)x=t as x→0,t→1 ∴t→1Lim​lntet−1−elnt​ ∴elntLimt→1​lntet−1−lnt−1​=t→1Lim​lntt−1−lnt​=t→1Lim​t1​1−t1​​=0.

Q. The value of $x \to 0^{+} Lim \frac{e ^{(s i n x)^{x} - 1} - e ^{x l n (s i n x)}}{x l n ( s i n x )}$ is equal to

$e$

$e - 1$

Let $(sin x)^{x} = t$
$as x \to 0, t \to 1$
$∴ t \to 1 Lim \frac{e ^{t - 1} - e ^{l n t}}{l n t}$
$∴ e^{l n t} Lim_{t \to 1} \frac{e ^{t - 1 - l n t} - 1}{l n t} = t \to 1 Lim \frac{t - 1 - l n t}{l n t} = t \to 1 Lim \frac{1 - \frac{1}{t}}{\frac{1}{t}} = 0.$