Finding Limits: limit x->0 (cos x - 1) / x^2, sin^2(x) / x^2

[SilvrNitr8]

I took Calculus a long time ago and I can't remember how to do limits. I was wondering if someone can show me step by step on how to take the limits of the following:

limit x->0 (cosx-1)/x^2

limit x-> sin^2(x)/x^2

Thanks

 

[soroban]

Re: Finding Limits

Hello, SilvrNitr8!

$\displaystyle \text{We're expected to know that: }\;\lim_{\theta\to0}\frac{\sin\theta}{\theta} \:=\:1$

I'll do the second one first . . .

$\displaystyle \lim_{x\to0}\frac{\sin^2\!x}{x^2}$

$\displaystyle \text{We have: }\;\lim_{x\to0}\left(\frac{\sin x}{x}\right)^2 \;=\;1^2 \;=\;1$

$\displaystyle \lim_{x\to0}\frac{\cos x-1}{x^2}$

$\displaystyle \text{Multiply top and bottom by }(\cos + 1)$

. . $\displaystyle \frac{\cos x - 1}{x^2}\cdot\frac{\cos x+ 1}{\cos x+ 1} \;=\; \frac{\cos^2\!x-1}{x^2(\cos x+1)} \;=\;\frac{-(1-\cos^2\!x)}{x^2(\cos x + 1)} \;=\;\frac{-\sin^2\!x}{x^2(\cos x+1)}$

$\displaystyle \text{Then: }\;\lim_{x\to0}\left(\frac{\sin x}{x}\right)^2\cdot\frac{-1}{\cos x + 1} \;=\;(1^2)\cdot\frac{-1}{1+1} \;=\;-\frac{1}{2}$