If f(x)=( sin (ex-2-1)/ log (x-1)), then displaystyle lim x arrow 2 f(x) is given by

Question

If f(x)=( sin (ex-2-1)/ log (x-1)), then displaystyle lim x arrow 2 f(x) is given by (A) -2 (B) -1 (C) 0 (D) 1

Tardigrade Admin · Accepted Answer

We have, displaystyle lim x arrow 2 f(x) = lim x arrow 2 ( sin (ex-2-1)/ log (x-1)) = displaystyle lim x arrow 2 ( sin (ex-2-1)/ex-2-1) ⋅ (ex-2-1/x-2) ⋅ (x-2/ log [1+(x-2)]) =1 × 1 × 1=1

Q. If $f (x) = \frac{s i n ( e ^{x - 2} - 1 )}{l o g ( x - 1 )}$ , then $x \to 2 lim f (x)$ is given by

-2

-1

0

1

We have,
$x \to 2 lim f (x) = x \to 2 lim \frac{sin ( e ^{x - 2} - 1 )}{lo g ( x - 1 )}$
$= x \to 2 lim {\frac{sin ( e ^{x - 2} - 1 )}{e ^{x - 2} - 1} \cdot \frac{e ^{x - 2} - 1}{x - 2} \cdot \frac{x - 2}{lo g [ 1 + ( x - 2 )]}}$
$= 1 \times 1 \times 1 = 1$

Q. If f(x)=log(x−1)sin(ex−2−1)​, then x→2lim​f(x) is given by

-2

-1

0

1

We have, x→2lim​f(x)=x→2lim​log(x−1)sin(ex−2−1)​ =x→2lim​{ex−2−1sin(ex−2−1)​⋅x−2ex−2−1​⋅log[1+(x−2)]x−2​} =1×1×1=1

Q. If $f (x) = \frac{s i n ( e ^{x - 2} - 1 )}{l o g ( x - 1 )}$ , then $x \to 2 lim f (x)$ is given by

We have,
$x \to 2 lim f (x) = x \to 2 lim \frac{sin ( e ^{x - 2} - 1 )}{lo g ( x - 1 )}$
$= x \to 2 lim {\frac{sin ( e ^{x - 2} - 1 )}{e ^{x - 2} - 1} \cdot \frac{e ^{x - 2} - 1}{x - 2} \cdot \frac{x - 2}{lo g [ 1 + ( x - 2 )]}}$
$= 1 \times 1 \times 1 = 1$