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Known facts.

$f$ and $g$ are continious and exist in the neighbourhood of $x=0$.

$\lim_{x\to 0}a(x)=\lim_{x\to 0}b(x)=0$.

$\lim_{x\to 0} \frac{f(x)}{g(x)}$ exists.

$\lim_{x\to 0} \frac{a(x)}{b(x)} = 1$


I want to investigate and possibly also prove the statement

$$\lim_{x\to 0}\frac{f\left(a(x)\right)}{g\left(b(x)\right)}=\lim_{x\to 0}\frac{f(x)}{g(x)}$$

My thoughts.

I know that $$\lim_{x\to 0} f\left(a(x)\right)=\lim_{x\to 0} f(x)=f(0)$$ and $$ \lim_{x\to 0} g(b(x))=\lim_{x\to 0} g(x)=g(0)$$

In the case of $g(0)\neq 0$, we get

$$\lim_{x\to 0} \frac{f(a(x))}{g(b(x))}=\frac{\lim_{x\to 0} f(a(x))}{\lim_{x\to 0} g(b(x))}=\frac{\lim_{x\to 0} f(x)}{\lim_{x\to 0} g(x)}=\lim_{x\to 0}\frac{f(x)}{g(x)}$$

but this is not valid when $g(0)=0$ (since it would give division by zero) and that may be the case here.

Now I really don't know how to proceed to prove t. I'm thinking that the $\epsilon-\delta$ defintions of the limits may be useful somehow, but I can't see how. This problem doesn't really seem to resemble anything I have seen before...

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Under the assumptions given, the desired conclusion is not true in general. Just let $f(x)=g(x)=a(x)=x$ and let $b(x)=2x$. Clearly $$\lim_{x\to0}a(x)=\lim_{x\to0}b(x)=0$$ but $$\lim_{x\to0}{f(x)\over g(x)}=1$$ while

$$\lim_{x\to0}{f(a(x))\over g(b(x))}={1\over2}$$

A suggestion: See what you can prove if you add the assumption

$$\lim_{x\to0}{a(x)\over b(x)}=1$$

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Let $f(x)=g(x)=x$, $a(x)=x$ and $b(x) = x^2$.

$\lim_{x\rightarrow 0}\frac{f(x)}{g(x)} = 1$, but $\lim_{x\rightarrow 0}\frac{f(a(x))}{g(b(x))} = \lim_{x\rightarrow 0}\frac{x}{x^2}$ does not exist.

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Other answers have given very simple and good counterexamples. I want to add an informal reasoning as to why the result fails under given assumptions. In the fraction $f(x)/g(x)$ the argument of both functions $f$ and $g$ is same i.e. $x$ and this argument tends to zero. This means that the argument of both $f$ and $g$ tend to zero is exactly the same manner. In the fraction $f(a(x))/g(b(x))$ the arguments for $f, g$ are $a(x), b(x)$ respectively and both these tend to zero, but given the arbitrary nature of $a(x), b(x)$ they may tend to zero in completely different manner. For example $a(x)$ may tend to zero much faster (or slower) compared to $b(x)$. If we ensure that $a(x)$ and $b(x)$ tend to zero with equal rate then the result can be established. This roughly means that $a(x) \approx b(x)$ and more formally $a(x)/b(x)$ should tend to $1$ as $x \to 0$.

You should follow Barry Cipra's suggestion to prove your result under the added assumption of $\lim_{x \to 0}\dfrac{a(x)}{b(x)} = 1$.

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