How to find the limit, without L'Hôpital's rule, of a function with exponentials?

Question

I have done some digging and I cannot find any posts addressing limits with exponentials and without L'Hôpital's rule.

I have one of these questions for my assignment, but for ethical reasons I have made up a similar function:

Find the following limit without L'Hôpital's rule: $$\lim_{x\to0}\frac{2^x-7^x}{2x}$$

Answer 1

Consider $$ \frac{2^x-7^x}{2x}=\frac{1}{2}\frac{2^x-1+1-7^x}{x}= \frac{1}{2}\left(\frac{2^x-1}{x}-\frac{7^x-1}{x}\right) $$ This is suggested by the fact we should know how to find the derivative of $f(x)=a^x$ and so we just need to compute the limit of the inner fraction (which should exist, in order for splitting the sum). So you can rewrite your limit as $$ \frac{1}{2}\left(\lim_{x\to0}\frac{2^x-1}{x}-\lim_{x\to0}\frac{7^x-1}{x}\right) $$ Now $$ \lim_{x\to0}\frac{a^x-1}{x}= \lim_{x\to0}\frac{e^{x\log a}-1}{x}\overset{(*)}{=} \lim_{t\to0}\frac{e^t-1}{t/\log a}=\log a $$ (log means "natural logarithm"). In the equality marked with $(*)$, the substitution $t=x\log a$ is used. Finally, the basic limit $$ \lim_{t\to0}\frac{e^t-1}{t}=1 $$ allows to conclude.

Answer 2

You can transform them to MacLaurins series f(x)=f(0)+f'(0)x/1!+f"(0)x²/2!+...

so $2^x=1+xln2+x^2ln^22/2+...$ and $7^x=1+xln7+x^2ln^27/2+...$

you have lim equal to (ln2-ln7+o(x))/2

(ln means "natural logarithm").

Answer 3

$$ \lim_{x\to 0}\frac{2^x-7^x}{2x} = \frac12\lim_{x\to 0}\frac{2^x-7^x-0}{x-0}$$ and the second limit is by definition the derivative of $x\mapsto 2^x-7^x$ at $x=0$. Differentiate this function symbolically and you're done.