$\lim_{x \to 0}\left(\frac{\sin^2(x)}{1-\cos(x)}\right)$ without L'Hopital's rule

Question

I'm trying to calculate $\lim_{x \to 0}\left(\frac{\sin^2(x)}{1-\cos(x)}\right)$ without L'Hopital's rule.

The trigonometrical identity $\sin^2(x) = \frac{1-\cos(2x)}{2}$ doesn't seem to lead anywhere. I also attempted to calculate using $\cos^2(x) + \sin^2(x) = 1$ without success.

Any ideas?

Answer 1

Consider the following:

$$\sin^2(x) = 1-\cos^2(x) = (1-\cos(x))(1+\cos(x))$$

You can do the cancellation and BABAM! Evaluating the limit after that is easy :D

Answer 2

Another way: $$\frac{\sin^2x}{1-\cos{x}}=2\cdot\dfrac{\dfrac{\sin^2x}{x^2}}{\dfrac{\sin^2\frac{x}{2}}{\dfrac{x^2}{4}}}\rightarrow2.$$

Answer 3

The way you tried to solve:

$(i) \ \displaystyle \sin^2(x) = \displaystyle \dfrac{1-\cos(2x)}{2}$

$$\displaystyle \lim _{x \rightarrow 0} \dfrac{1}{2}\left(\dfrac{1-\cos (2 x)}{1-\cos (x)}\right)$$

$$=\dfrac{1}{2} \displaystyle \lim _{x \rightarrow 0} \frac{\cos (0)-\cos (2 x)}{\cos(0)-\cos (x)}=\frac{1}{2} \displaystyle \lim _{x \rightarrow 0} \dfrac{\sin ^{2} x}{\sin ^{2} \frac{x}{2}}$$ (Using the identity $\cos A-\cos B=-2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)$ )

Using $(ii) \ \sin^2(\theta)+ \cos^2(\theta)=1$

$$\displaystyle \lim _{x \rightarrow 0} \dfrac{\sin ^{2} x}{1-\sin \left(\frac{\pi}{2}-x\right)}$$

Let $\frac{\pi}{2}-x=2\theta$

$$\displaystyle \lim _{x \rightarrow 0} \dfrac{\sin ^{2} x}{\cos ^{2} (\theta)+\sin ^{2}( \theta)-2 \sin (\theta) \cos (\theta)} =\displaystyle \lim _{x \rightarrow 0} \frac{\sin ^{2} x}{(\cos \theta-\sin \theta)^{2}} $$

$$\displaystyle \lim _{x \rightarrow 0} \dfrac{\sin^2(x)}{\left[\cos \left(\frac{\pi}{4}-\frac{x}{2}\right)-\sin \left(\frac{\pi}{4}-\frac{x}{2}\right)\right]^{2}}$$

Expand the sine and cosine terms with respective formula to again get: $$\displaystyle \lim _{x \rightarrow 0} \dfrac{\sin ^{2} x}{2\sin ^{2} \frac{x}{2}}$$