Limx - 0x2cos x1 - cos x is

The correct option is A 2 Using L hospital rule =x^2( sinx)+cosx(2x)sinx again using L hospital rule = x^2cosx 2xsinx 2xsinx+2cosxcosx now putting ...

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

Byju's Answer

Standard XII

Mathematics

Monotonically Increasing Functions

limx → 0x2cos...

Question

$lim x \to 0 \frac{x^{2} cos x}{1 - cos x}$ is

2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{3}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{- 3}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $2$
Using L-hospital rule
$= \frac{x^{2} (- s i n x) + c o s x (2 x)}{s i n x}$
again using L hospital rule
$= \frac{- x^{2} c o s x - 2 x s i n x - 2 x s i n x + 2 c o s x}{c o s x}$
now putting limit $x \to 0$
$= 2$

Suggest Corrections

Similar questions

Q. Solve

lim x \to 0 \frac{x^{2} cos x}{1 - cos x}

\lim_{x \to 1} {\{\frac{x^{3} + 2 x^{2} + x + 1}{x^{2} + 2 x + 3}\}}^{\frac{1 - \cos (x - 1)}{{(x - 1)}^{2}}}

lim x \to 0 \frac{1 - cos x}{x^{2}} [\frac{1}{2}]

l i m x \to \frac{π}{2} [\frac{[s i n x] - [c o s x] + 1}{3}] =

[.] \to

denotes greatest integer function

Q. Evaluate:

lim x \to 0 \frac{s i n^{2} x}{\sqrt{3 + c o s x} - 2}

Join BYJU'S Learning Program

limx→0x2cosx1−cosx is

The correct option is A 2Using L-hospital rule=x2(−sinx)+cosx(2x)sinxagain using L hospital rule=−x2cosx−2xsinx−2xsinx+2cosxcosxnow putting limit x→0=2

$lim x \to 0 \frac{x^{2} cos x}{1 - cos x}$ is

The correct option is A $2$
Using L-hospital rule
$= \frac{x^{2} (- s i n x) + c o s x (2 x)}{s i n x}$
again using L hospital rule
$= \frac{- x^{2} c o s x - 2 x s i n x - 2 x s i n x + 2 c o s x}{c o s x}$
now putting limit $x \to 0$
$= 2$