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If $\lim_{ x\to 0} f(x^2) = L$, is the $\lim_{ x\to 0} f(x)$ also equal to $L$? If not, prove by an example.

Firstly, I apoligise for the horrible formatting, it doesn't seem to be working on my computer.

Secondly, I am having a hard time figuring this out. It seems to me that you could (via a proof) but I can think of a few examples where the limit is not equal, for example $x^{\frac{1}{2}}$.

Any thoughts on how to approach this?

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    Note that under the conditions of the question we do have $\lim_{x\to 0^{+}}f(x)=L$. – 2017-02-08

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No. For example $f(x)=1$ if $x>=0$ and 0 otherwise. Then $\lim_{x\rightarrow 0}f(x^2)=1$, yet $\lim_{x\rightarrow 0}f(x)$ is not defined.

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    All to easy. (+1) – 2017-02-08
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    Is there any other example apart from that one? – 2017-02-08
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    @JDel1996 Do you understand the concept from Alex's example? Because if so, it should be easy to construct infinitely many examples. – 2017-02-08
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No, another example is $f(x^2)=e^{\frac{-1}{x^2}}$, note that $x^2$ is always positive where $x$ is not. Using this you can find infinitely many examples.

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    That doesn't work. Did you mean to write $f(x^2)$? – 2017-02-08
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    oh! sorry for that ! yeah I mean $f(x^2)$ – 2017-02-08
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Another counterexample:

Let $f(x) = c$ for $x \ge 0$ and $f(x) = -c$ for $x < 0$ where $c > 0$.

Then $f(x^2) = c$ for all real $x$ but $\lim_{x \to 0^-} f(x) =-c$ and $\lim_{x \to 0^+} f(x) =c$.