$ \lim_{x\to 0} \frac{\cos(x)}{x^2} $

Question

$$ \lim_{x\to 0} \frac{\cos(x)}{x^2} $$

How do I solve this without l'Hospital? Online calculator gives me infinity, but I'm sure there's an answer because I've solved several of these algebraically when the calculators give me indeterminate/infinity etc

Also tips on how to solve these trig limit problems would be helpful so I can be independent.

Answer 1

No, the limit is $+\infty$, since $lim_{x\rightarrow 0}cosx=1$, $lim_{x\rightarrow 0}x^2=0$ and $x^2>0$.

Answer 2

To prove that $\lim_{x\to0}f(x)=+\infty$ it must be shown that for every $m>0$ you can find a $\delta_m>0$ such that $$0<|x|\leq\delta_m\implies f(x)>m$$

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Let $m>0$.

For some $c>0$ we will have $|x|<c\implies\cos x>\frac12$.

This as a consequence of $\cos0=1>\frac12$ and the fact that the function is continuous.

$0<|x|<\delta_m:=\min\left(c,\frac1{\sqrt{2m}}\right)$.

Then we have: $$0<|x|<\delta_m:=\min\left(c,\frac1{\sqrt{2m}}\right)\implies\frac{cosx}{x^2}>\frac{\frac12}{\frac1{2m}}=m$$

This proves that: $$\lim_{x\to0}\frac{cosx}{x^2}=+\infty$$