Let f(x)= begincases((1+ tan x)(1/x)-e/x), text if x ≠ 0 k, text if x=0 endcases If f ( x ) is continuous at x=0, then the value of k is

Question

Let f(x)= begincases((1+ tan x)(1/x)-e/x), text if x ≠ 0 k, text if x=0 endcases If f ( x ) is continuous at x=0, then the value of k is (A) e (B) (-e/2) (C) (-e/4) (D) None

Tardigrade Admin · Accepted Answer

k=e undersetx arrow 0 textLim (e( ln (1+ tan x)-1/x)-1/x) = e undersetx arrow 0 textLim ( ln (1+ tan x)-x/x2) k=e undersetx arrow 0 textLim ( tan x-( tan 2 x/2)+ ldots ldots ldots-x/x2)=(-e/2) .

Q. Let $f (x) = {\frac{( 1 + t a n x ) ^{\frac{1}{x}} - e}{x}, k, if x \neq = 0 if x = 0$ If $f (x)$ is continuous at $x = 0$ , then the value of $k$ is

$e$

$\frac{- e}{2}$

$\frac{- e}{4}$

None

$k = e x \to 0 Lim \frac{e ^{\frac{l n ( 1 + t a n x ) - 1}{x} - 1}}{x}$
$= e x \to 0 Lim \frac{l n ( 1 + t a n x ) - x}{x ^{2}}$
$k = e x \to 0 Lim \frac{t a n x - \frac{t a n ^{2} x}{2} + \dots\dots\dots - x}{x ^{2}} = \frac{- e}{2} .$

Q. Let f(x)={x(1+tanx)x1​−e​,k,​ if x=0 if x=0​ If f(x) is continuous at x=0, then the value of k is

e

2−e​

4−e​

None

k=ex→0Lim​xexln(1+tanx)−1​−1​ =ex→0Lim​x2ln(1+tanx)−x​ k=ex→0Lim​x2tanx−2tan2x​+………−x​=2−e​.

Q. Let $f (x) = {\frac{( 1 + t a n x ) ^{\frac{1}{x}} - e}{x}, k, if x \neq = 0 if x = 0$ If $f (x)$ is continuous at $x = 0$ , then the value of $k$ is

$e$

$\frac{- e}{2}$

$\frac{- e}{4}$

$k = e x \to 0 Lim \frac{e ^{\frac{l n ( 1 + t a n x ) - 1}{x} - 1}}{x}$
$= e x \to 0 Lim \frac{l n ( 1 + t a n x ) - x}{x ^{2}}$
$k = e x \to 0 Lim \frac{t a n x - \frac{t a n ^{2} x}{2} + \dots\dots\dots - x}{x ^{2}} = \frac{- e}{2} .$