displaystyle limx→0 (1+tan x/1+sin x) cosec x is equal to

Question

displaystyle limx→0 (1+tan x/1+sin x) cosec x is equal to (A)  (1/e)  (B)  1  (C)  e  (D)  e2

Tardigrade Admin · Accepted Answer

lim x arrow 0 (1+ tan x/1+ sin x) operatornamecosec x = lim x arrow 0 ([.(1+( sin x/ cos x))( cos x/ sin x)|1 / cos x./(1+ sin x)1 / sin x) =(e lim x arrow 0 (1/ cos x)/e)=(e/e)=1

Q. $x \to 0 lim {\frac{1 + t an x}{1 + s in x}}^{cosec x}$ is equal to

$\frac{1}{e}$

$1$

$e$

$e^{2}$

$lim_{x \to 0} {\frac{1 + t a n x}{1 + s i n x}}^{cosec x}$
$= lim_{x \to 0} \frac{[ ( 1 + \frac{s i n x}{c o s x} ) ^{\frac{c o s x}{s i n x}} ∣ ∣ ^{1/ c o s x}}{( 1 + s i n x ) ^{1/ s i n x}}$
$= \frac{e ^{l i m_{x \to 0} \frac{1}{c o s x}}}{e} = \frac{e}{e} = 1$

Q. x→0lim​{1+sinx1+tanx​}cosecx is equal to

e1​

1

e

e2

limx→0​{1+sinx1+tanx​}cosecx =limx→0​(1+sinx)1/sinx[(1+cosxsinx​)sinxcosx​∣∣​1/cosx​ =eelimx→0​cosx1​​=ee​=1

Q. $x \to 0 lim {\frac{1 + t an x}{1 + s in x}}^{cosec x}$ is equal to

$\frac{1}{e}$

$1$

$e$

$e^{2}$

$lim_{x \to 0} {\frac{1 + t a n x}{1 + s i n x}}^{cosec x}$
$= lim_{x \to 0} \frac{[ ( 1 + \frac{s i n x}{c o s x} ) ^{\frac{c o s x}{s i n x}} ∣ ∣ ^{1/ c o s x}}{( 1 + s i n x ) ^{1/ s i n x}}$
$= \frac{e ^{l i m_{x \to 0} \frac{1}{c o s x}}}{e} = \frac{e}{e} = 1$