Compute the limit $\lim_{x\to 0} \frac{2^x+5^x-4^x-3^x}{5^x+4^x-3^x-2^x}$

Question

I don't know how to start, i've noticed that it can be written as $$\lim_{x\to 0} \frac{2^x+5^x-4^x-3^x}{5^x+4^x-3^x-2^x}=\lim_{x\to 0} \frac{(5^x-3^x)+(2^x-4^x)}{(5^x-3^x)-(2^x-4^x)}$$

Answer 1

$$\lim_{x\rightarrow0}\frac{2^x+5^x-4^x-3^x}{5^x+4^x-3^x-2^x}=\lim_{x\rightarrow0}\frac{\frac{2^x-1}{x}+\frac{5^x-1}{x}-\frac{4^x-1}{x}-\frac{3^x-1}{x}}{\frac{5^x-1}{x}+\frac{4^x-1}{x}-\frac{3^x-1}{x}-\frac{2^x-1}{x}}=$$ $$=\frac{\ln2+\ln5-\ln4-\ln3}{\ln5+\ln4-\ln3-\ln2}.$$ I used $$\lim_{x\rightarrow0}\frac{a^x-1}{x}=\ln{a}\lim_{x\rightarrow0}\frac{e^{x\ln{a}}-1}{x\ln{a}}=\ln{a}$$ for $a>0$.

Answer 2

This is a very special case of l’Hôpital’s theorem, possibly what gave him (or Bernoulli) the idea for the general case.

If you have two functions $f$ and $g$ which are differentiable at $0$ (or any other point), then $$ \lim_{x\to0}\frac{f(x)-f(0)}{g(x)-g(0)}=\frac{f'(0)}{g'(0)} $$ provided $g'(0)\ne0$ (if $g'(0)=0$, it's another story). The link with the general theorem should be apparent.

However, this is not l’Hôpital: the proof is very easy, because one notices that $$ \lim_{x\to0}\frac{f(x)-f(0)}{g(x)-g(0)}= \lim_{x\to0}\frac{\,\dfrac{f(x)-f(0)}{x}\,}{\,\dfrac{g(x)-g(0)}{x}\,}= \frac{f'(0)}{g'(0)} $$ by standard limit theorems.

Thus your limit boils down to considering $f(x)=2^x+5^x-4^x-3^x$ and $g(x)=5^x+4^x-3^x-2^x$, since $f(0)=0$ and $g(0)=0$. Then the limit is $$ \frac{\log2+\log5-\log4-\log3}{\log5+\log4-\log3-\log2} $$

But I can’t use derivatives! you might object. Well, it happens that taking the derivative is considered an application of l’Hôpital, but actually it isn’t, as shown before. I use to say that when we know derivatives we are able to compute limits of several sorts, so why avoiding them?

If you can't use derivatives, just pretend you don't. Let's look at $f$; it is a sum of functions. How do you prove that the derivative of a sum is the sum of the derivatives?

Suppose you have $a(x)$ and $b(x)$, with $s(x)=a(x)+b(x)$ (always at $0$, but it's the same at every point): $$ \frac{s(x)-s(0)}{x-0}=\frac{(a(x)-a(0))+(b(x)-b(0))}{x}= \frac{a(x)-a(0)}{x}+\frac{b(x)-b(0)}{x} $$ If the function are four, say $a$, $b$, $c$ and $d$, with $s=a+b+c+d$, then $$ \frac{s(x)-s(0)}{x-0}= \frac{a(x)-a(0)}{x}+\frac{b(x)-b(0)}{x}+ \frac{c(x)-c(0)}{x}+\frac{d(x)-d(0)}{x} $$ Now you see how to hide the derivative under the carpet: $$ \lim_{x\to0}\frac{2^x+5^x-4^x-3^x}{5^x+4^x-3^x-2^x} = \lim_{x\to0}\frac{\dfrac{2^x-1}{x}+\dfrac{5^x-1}{x}-\dfrac{4^x-1}{x}-\dfrac{3^x-1}{x}}{\dfrac{5^x-1}{x}+\dfrac{4^x-1}{x}-\dfrac{3^x-1}{x}-\dfrac{2^x-1}{x}} $$

You know that $$ \lim_{x\to0}\frac{r^x-1}{x}=\log r $$ is the derivative at $0$ of $x\mapsto r^x$, don't you?