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: $$\lim_{x\to0}\frac{2^x-7^x}{2x}$$

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    $\begingroup$ Did you study Taylor series? Most of the time this kind of limit can be solved by expanding the numerator/denominator using Taylor... (I had a physics professor that always said they should ban teachin de L'Hopital and instead teach Taylor series as a fundamental operation). $\endgroup$ – Bakuriu Sep 28 '16 at 18:10
  • $\begingroup$ No, we haven't looked at Taylor series $\endgroup$ – Skylineblue Sep 29 '16 at 6:18

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.

  • $\begingroup$ Thanks Egreg. Can you explain the rewriting for me please? $\endgroup$ – Skylineblue Sep 28 '16 at 12:22
  • $\begingroup$ and also how you got to (e^t -1)/(t/log a)? and how this equals log a? $\endgroup$ – Skylineblue Sep 28 '16 at 12:24
  • $\begingroup$ $$ \lim_{x\to0}\frac{a^x-1}{x}= \lim_{x\to0}\frac{e^{x\log a}-1}{x}= \lim_{t\to0}\frac{e^t-1}{t/\log a}=\log a $$ can be taken as formula, right? $\endgroup$ – Fawad Sep 28 '16 at 12:25
  • $\begingroup$ Can you justify lim x -> 0 (e^(xlog a) -1)/x = log a through the identify lim x -> 0 (e^x -1)/x = 1? $\endgroup$ – Skylineblue Sep 28 '16 at 12:52

$$ \lim_{x\to 0}\frac{2^x-7^x}{2x} = \frac12\lim_{x\to 0}\frac{2^x-7^x-0}{x-0}$$ 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.

  • $\begingroup$ Thanks for your response Henning, but I didn't quite understand your method (and thus have chosen Egreg's response). How is the second limit by definition the derivative of x↦2^x -7^x at x=0? Also, what do you mean by differentiating symbolically? $\endgroup$ – Skylineblue Sep 29 '16 at 9:03
  • $\begingroup$ @Skylineblue: The definition of the derivative of $f$ at $a$ is $\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$, and the limit above has this form for $a=0$. Differentiating symbolically means using calculus rules to find an expression for $f'(x)$. Once you have done this, you can plug $x=0$ into that expression. $\endgroup$ – hmakholm left over Monica Sep 29 '16 at 9:10
  • $\begingroup$ Thanks for the clarification. $\endgroup$ – Skylineblue Sep 29 '16 at 9:40

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").


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