Question

Lim x tends to zero 1 - cos x upon sin square x​

Answer 1

Answer:

The answer is $\dfrac{1}{2}$

Step-by-step explanation:

 $\displaystyle \lim_{x\to 0}\dfrac{1-\cos x }{\sin^2x}$

Rationalize numerator and simplify

$=\displaystyle \lim_{x\to 0}\dfrac{1-\cos x }{\sin^2x}\times \dfrac{1+\cos x}{1+\cos x}\\\\\\=\displaystyle \lim_{x\to 0}\dfrac{1-\cos^2 x }{\sin^2x}\times \dfrac{1}{1+\cos x}\\\\\\=\displaystyle \lim_{x\to 0}\dfrac{\sin^2x }{\sin^2x}\times \dfrac{1}{1+\cos x}\\\\\\=\displaystyle \lim_{x\to 0}\dfrac{1}{1+\cos x}=\dfrac{1}{1+\cos 0}=\dfrac{1}{1+1}=\dfrac{1}{2}$