Evaluate $\lim_{x\to 0} \frac{a^x -1}{x}$ without applying L'Hopital's Rule. The questions is: Evaluate $$\lim_{x\to 0} \frac{a^x -1}{x}$$ without applying L'Hopital's Rule.
Does this question fundamentally same as asking if the $\lim_{x\to 0} \frac{a^x -1}{x}$ exists? rather than straightway asking to find the limit. That means are questions (1) proving if the limit of a function exists and (2) asking what is the limit of that function, essentially same question?
 A: we have
$$a^x=e^{x\ln(a)} $$
and
$$\lim_{t\to 0}\frac {e^t-1}{ t} =1$$
the limit is $$\ln(a) $$
A: set $$t=a^x-1$$ then we have
$$x=\frac{1}{\ln(a)}\ln(t+1)$$ and you will get
$$\frac{t}{\frac{1}{\ln(a)}\ln(t+1)}$$
can you finish?
A: Another way would be by Taylor expanding $$a^x = e^{\ln(a)x} = 1 + \ln(a)x + O(x^2)$$
And we get $$\frac{1+\ln(a)x - 1 + O(x^2)}{x} = \ln(a) + O(x) \to \ln(a)\text{ as } x \to 0$$
A: Hint: if you write $f(x) = a^x$, then the given limit has the form $$\lim_{x \to 0} \frac{f(x) - f(0)}{x-0}.$$ Does that look familiar?
A: Evaluating the limit $\mathop {\lim }\limits_{x \to 0} \frac{{{a^x} - 1}}{x}$ without applying the L'Hospital's Rule.
I think that one of best way for evaluating of this limit is using of the power series expansion for the function $a^x, (a > 0)$ about the point $x = 0$. Namely, we can write that
$$
\begin{align*}
a^x &= \sum_{k = 0}^\infty  \left( x \ln(a) \right)^k  \\
&= 1 + x \ln(a) + x^2 \ln(a)^2 + x^3 \ln(a)^3 +  \cdots  \\
&= 1 + x \ln(a) \left( 1 + x \ln(a) + x^2 \ln(a)^2 + x^3 \ln(a)^3 +  \cdots \right) \\
&= 1 + x \ln(a) \sum_{k = 0}^\infty  \left( x \ln(a) \right)^k \\
&= 1 + x \ln(a) \lim_{n \to \infty} \sum_{k = 0}^n \left( x \ln(a) \right)^k \\
&= 1 + x \ln(a) \lim_{n \to \infty } \frac{1 - \left(x \ln(a) \right)^{n + 1}}{1 - x \ln(a)}
\end{align*}
$$
by the sum of the geometric sequence.
So, we have got
$$a^x = 1 + x \ln(a) \lim_{n \to \infty } \frac{1 - \left( x \ln(a) \right)^{n + 1}}{1 - x \ln(a)}.$$
Substitution of this expression in $\lim_{x \to 0} \frac{a^x - 1}{x}$ gives us the following:
$$
\begin{align*}
\lim_{x \to 0} \frac{a^x - 1}{x} &= \lim_{x \to 0} \frac{1 + x \ln(a) \lim_{n \to \infty} \frac{1 - \left( x \ln(a) \right)^{n + 1}}{1 - x \ln(a)} - 1}{x} \\
&= \lim_{x \to 0} \frac{x \ln(a) \lim_{n \to \infty } \frac{1 - \left( x \ln(a) \right)^{n + 1}}{1 - x \ln(a)}}{x}  \\
&= \ln(a) \lim_{x \to 0} \left( \lim_{n \to \infty } \frac{1 - \left( { \ln(a)} \right)^{n + 1}}{1 - x \ln(a)} \right) \\
&= \ln(a) \lim_{n \to \infty } \left( \lim_{x \to 0} \frac{1 - \left( x \ln(a) \right)^{n + 1}}{1 - x \ln(a)} \right) \\
&= \ln(a) \frac{1 - 0}{1 - 0} \\
&= \ln(a).
\end{align*}
$$
So, $\mathop {\lim }\limits_{x \to 0} \frac{{{a^x} - 1}}{x}= \ln(a)$.
