Question

lim x tends to 0 1-cosx /sin^2x

Answer 1

Please find the attached image of solution.

Answer 2

Answer:  The required value of the given limit is $\dfrac{1}{2}.$

Step-by-step explanation:  We are given to find the value of the following limit :

$\lim_{x \rightarrow 0}\dfrac{1-\cos x}{\sin^2x}.$

Applying L'Hospital's rule, we have

$\lim_{x \rightarrow 0}\dfrac{1-\cos x}{\sin^2x}~~~[\frac{0}{0}]\\\\\\=\lim_{x \rightarrow 0}\dfrac{\frac{d}{dx}(1-\cos x)}{\frac{d}{dx}(\sin^2x)}\\\\\\=\lim_{x \rightarrow 0}\dfrac{\sin x}{2\sin x\cos x}\\\\\\=\lim_{x\rightarrow 0}\dfrac{1}{2\cos x}\\\\\\=\dfrac{1}{2\cos 0}\\\\\\=\dfrac{1}{2\times1}\\\\\\=\dfrac{1}{2}.$

Thus, the required value of the given limit is $\dfrac{1}{2}.$