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...


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

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


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.


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.