Help evaluating $\lim_{x \to 0}{\frac{a^{k} - a^{k - x}}{bx}}$ without L'Hospital?

Question

Evaluate $\displaystyle\lim_{x \to 0}{\frac{a^{k} - a^{k - x}}{bx}}$ if $a$, $b$, and $k$ are positive real numbers.

For this attempt, let $L$ be the limit, if it exists. Then, \begin{align*}L &\;=\; \lim_{x \to 0}{\frac{(a^{k - x})(a^{x} - 1)}{bx}} \\ L &\;=\; \left[\lim_{x \to 0}a^{k - x}\right]\left[\lim_{x \to 0}\frac{a^{x} - 1}{b x}\right]\end{align*}

The first factor approaches $a^{k}$ and the second factor approaches $\frac{0}{0}$. Is there a way to get the limit of the second factor without L'Hospital's rule?

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Edit: I just found a similar problem which is $\displaystyle \lim_{x \to 0}\frac{e^{x} - 1}{x}$ using table of values. Is there an algebraic way to solve this?

Answer 1

So you are left with computing the limit $$ \lim_{x\to 0} \frac{a^x-1}{bx}\,. $$ Since $b>0$ is a constant, it suffices to find $$ \lim_{x\to 0} \frac{a^x-1}{x} = \lim_{x\to 0} \frac{a^x-a^0}{x-0}\,, $$ where we rewrote it in that way (on the RHS) to recognize a derivative at $0$. Namely, letting $f\colon \mathbb{R}\to\mathbb{R}$ defined by $f(x) = a^x=e^{x\ln a}$, we have $$ f'(0) = \lim_{x\to 0} \frac{f(x)-f(0)}{x-0} = \lim_{x\to 0} \frac{a^x-a^0}{x-0} $$ which is the quantity we seek to determine. So all you need is the derivative of $f$ at $0$ to conclude. But the derivative of $f$ is just $f'(x) = (\ln a) e^{x\ln a}$ (chain rule+derivative of the exponential), so $\boxed{f'(0) = \ln a}$.

Answer 2

You can pull out irrelevant constants with

$$\frac{a^{k}-a^{k-x}}{bx}=\frac{a^k}{b}\frac{1-e^{-x\log a}}{x}=\log a\frac{a^k}{b}\frac{1-e^{-x\log a}}{x\log a}.$$ The last factor is well-known to tend to $1$.