Limx- 0xex-1-2cos x -1-x1-cosx-

Using L' Hospital Rule Req. Limit = lim_x→ 0(e^x 1)+xe^x 2sin x /(1 cosx)+x(sinx) ( 0/0 ) Again using L' Hospital rule = lim_x→ 0(e^x+e^x+xe^x 2cosx)/sinx+si ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

First Principle of Differentiation

limx→ 0xex-1+...

Question

${lim}_{x \to 0} \frac{x (e^{x} - 1) + 2 (c o s x - 1)}{x (1 - c o s x)} =$
1

Open in App

Solution

The correct option is A 1
Using L' Hospital Rule

Req. Limit = ${lim}_{x \to 0} \frac{(e^{x} - 1) + x e^{x} - 2 s i n x}{(1 - c o s x) + x (s i n x)} (\frac{0}{0})$

Again using L' Hospital rule

$= {lim}_{x \to 0} \frac{(e^{x} + e^{x} + x e^{x} - 2 c o s x)}{s i n x + s i n x + x c o s x} (\frac{0}{0})$

Again using L' Hospital Rule

= ${lim}_{x \to 0} \frac{e^{x} + e^{x} + e^{x} + x e^{x} + 2 s i n x}{c o s x + c o s x + c o s x - x s i n x}$

= $\frac{1 + 1 + 1 + 0 + 0}{1 + 1 + 1 - 0} = 1$

Suggest Corrections

Similar questions

{lim}_{x \to 0} \frac{x (e^{x} - 1)}{1 - c o s x}

Q. Solve

lim x \to 0 \frac{x (e^{x} - 1)}{1 - cos x}

lim x \to 0 \frac{x (e^{x} - 1)}{1 - cos x}

is equal to

$lim_{x\to 0}$ $\frac{x({e}^{x}-1)}{1-cos x}$ =

\lim_{x \to 0} \frac{x (e^{x} - 1)}{1 - \cos x}

Join BYJU'S Learning Program

limx→0x(ex−1)+2(cosx−1)x(1−cosx)=1

The correct option is A 1Using L' Hospital Rule Req. Limit = limx→0(ex−1)+xex−2sinx(1−cosx)+x(sinx)(00) Again using L' Hospital rule =limx→0(ex+ex+xex−2cosx)sinx+sinx+xcosx(00) Again using L' Hospital Rule = limx→0ex+ex+ex+xex+2sinxcosx+cosx+cosx−xsinx = 1+1+1+0+01+1+1−0=1

${lim}_{x \to 0} \frac{x (e^{x} - 1) + 2 (c o s x - 1)}{x (1 - c o s x)} =$
1