Limit of $x[3-\cos(x^2)]$ using the epsilon-delta definition of a limit

Asked Mar 05 '17 at 22:12

Active Mar 03 '18 at 17:53

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I have been asked to determine the limit $L$ of $x[3-\cos(x^2)]$ for $a=0$, and prove that it is the limit by showing how to find a $\delta$ such that $|f(x)-L|< \epsilon$, $\forall x$ satisfying $0<|x-a|<\delta$.

I understand that by definition $\delta=\min(1,...)$ and that $\delta$ depends on our $\epsilon$.

Thanks!

edited Mar 03 '18 at 17:53

Lorenzo B.

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asked Mar 05 '17 at 22:12

Jose

So for clarification purposes you are trying to find $\lim_{x\to a}f(x)=L$ such that $a=0$ and $f(x):=3x-x\cos(x^2)$? – Sentinel135 Mar 05 '17 at 22:24
@Sentinel135 Yes- – Jose Mar 05 '17 at 22:26
Have you figured out what $L$ is? – Sentinel135 Mar 05 '17 at 22:27
Ok hint: $L=0$ and set $\delta<\varepsilon/4$. Also Keep in mind that $\cos(x^2)\in [-1,1],; \forall x \in \mathbb R$. – Sentinel135 Mar 05 '17 at 22:43

Limit of $x[3-\cos(x^2)]$ using the epsilon-delta definition of a limit

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