"Find the limit limx→0xsin⁡x2−2cos⁡x" solved and explained

Tazmin Horton

Answered question

2020-10-28

Find the limit

lim_{x \to 0} \frac{x \sin x}{2 - 2 \cos x}

likvau

Skilled2020-10-29Added 75 answers

To find the limit,

lim_{x \to 0} \frac{x \sin x}{2 - 2 \cos x}

Let

l = lim_{x \to 0} \frac{x \sin x}{2 - 2 \cos x}

\Rightarrow l = lim_{x \to 0} \frac{x \sin x}{2 (1 - \cos x)}

\Rightarrow l = lim_{x \to 0} \frac{x (2 \sin \frac{x}{2} \cos \frac{x}{2})}{2 (2 \sin^{2} \frac{x}{2})} ∵ \sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}, \cos x = 1 - 2 \sin^{2} \frac{x}{2}

\Rightarrow l = lim_{x \to 0} \frac{x \cos \frac{x}{2}}{2 \sin \frac{x}{2}}

\Rightarrow l = lim_{x \to 0} \frac{\frac{x}{2}}{\sin \frac{x}{2}} lim_{x \to 0} \cos \frac{x}{2}

\Rightarrow l = lim_{y \to 0} \frac{y}{\sin y} \cos 0 l e t \frac{x}{2} = y, t h e n a s x \to 0, y \to 0

\Rightarrow l = lim_{y \to 0} \frac{1}{\cos y}

(Applying LHospitals Rule)

\Rightarrow l = \frac{1}{\cos 0}

\Rightarrow l = 1

Therefore,

lim_{x \to 0} \frac{x \sin x}{2 - 2 \cos x} = 1

Jeffrey Jordon

Expert2022-08-24Added 2605 answers

Answer is given below (on video)

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?