# Limit problem involving trig functions and radical

I am stuck on this limit and have no idea how to solve it and which trig identity to use. Any help would be appreciated. Thanks!

$\lim\limits_{x \to 0^-} \frac{\sqrt{1+2\sin^2 \frac{x}{2}-\cos^2x}}{\left\lvert x \right\rvert}$

Note: Without using L'Hopitals rule.

• Do you know the answer? – mint Jan 24 '16 at 23:32
• Yeah, i think i finally got it. – Oliver Nikolovski Jan 25 '16 at 0:21

Hint: Using the identity $1-\cos^2x=\sin^2x$ one gets $$\lim\limits_{x \to 0^-} \frac{\sqrt{1+2\sin^2 \frac{x}{2}-\cos^2x}}{\left\lvert x \right\rvert}=\lim\limits_{x \to 0^-} \frac{\sqrt{\sin^2x+2\sin^2\tfrac x2}}{\sqrt{x^2}}=\lim\limits_{x \to 0^-}\sqrt{\dfrac{\sin^2x+2\sin^2\tfrac x2}{x^2}},$$ which can be easily computed by recalling that $\lim\limits_{u\to0}\dfrac{\sin u}{u}=1$.
Write the first few terms of the Taylor series of what is under the square root, which gives $1+2\sin^2\frac x 2-\cos^2 x=\frac32 x^2+\cdots$. Hence the limit is $\sqrt{3/2}$.