Question

limit x tends to 0 √(1-cosx)/x​

Answer 1

Given Expression,

$\displaystyle \: \sf \ \: \lim_{x \rightarrow \: 0} \: \dfrac{ \sqrt{1 - cos \: x} }{x}$

At 0, the above expression tends to 0/0 form.

We know that,

sin²∅ + cos²∅ = 1

cos2∅ = cos²∅ - sin²∅

Thus,

$\displaystyle \: \implies \sf \ \: \lim_{x \rightarrow \: 0} \: \dfrac{ \sqrt{ {sin}^{2} \dfrac{x}{2} + {cos}^{2} \dfrac{x}{2} - cos {}^{2} \: \dfrac{x}{2} + sin {}^{2} \dfrac{x}{2}} }{x} \\ \\ \displaystyle \implies \sf \ \: \lim_{x \rightarrow \: 0} \: \dfrac{ \sqrt{ {2sin}^{2} \dfrac{x}{2} } }{x} \\ \\\displaystyle \implies \sf \ \sqrt{2 } \ \times \lim_{x \rightarrow \: 0} \: \dfrac{{sin}^{} \dfrac{x}{2}}{x}$

We know that,

$\displaystyle \sf lim_{x \rightarrow 0} \dfrac{sin \ x}{x} = 1$

Thus,

$\displaystyle \implies \sf \ \sqrt{2 } \ \times \lim_{x \rightarrow \: 0} \: \dfrac{{sin} \dfrac{x}{2}}{2 \times \dfrac{x}{2} } \\ \\ \displaystyle \implies \sf \ \dfrac{\sqrt{2 }}{2} \ \times \lim_{x \rightarrow \: 0} \: \dfrac{{sin} \dfrac{x}{2}}{ \dfrac{x}{2} } \\ \\ \implies \sf \dfrac{1}{ \sqrt{2} }$