Finding $\lim\limits_{x\to 0} \frac{a^x-1}{x}$ without L'Hopital and series expansion.

Question

So we have to find $\lim\limits_{x\to 0} \frac{a^x-1}{x}$ without using any series expansions or the L'Hopital's rule.
I did it using both but I have no idea how to do it. I tried many substitutions but nothing worked. Please point me in the right direction.

Answer 1

$$x=\log_a(t)$$

$$\lim_{t\rightarrow 1}\frac{t-1}{\log_at}=\lim_{t\rightarrow 1}\frac{(t-1)\log(a)}{\log(t)}=\log(a)$$ EDIT:

$$L=\lim_{x\rightarrow 1}\frac{\log(x)}{x-1}=\lim_{x\rightarrow 0}\frac{\log(x+1)}{x}=\lim_{x\rightarrow 0}\int_0^x \frac 1x\frac{dt}{1+t}$$ Put: $t=xs$ then: $$L=\lim_{x\rightarrow 0}\int_0^1 \frac{ds}{1+xs}=\int_0^1ds=1$$

Also: $$L=\lim_{x\rightarrow 0}\frac{\log(1+x)-\log(1)}{(x+1)-1}=(\log(x))'_{x=1}=1$$

Answer 2

Let's try with substitution $x=\frac{t}{\log a}$

Using $a=e^{\log a}$ you obtain $$\lim_{t\to \infty} \left(\frac{a^{\frac{t}{\log a}}-1}{t}\right) \log a= \lim_{t\to \infty} \left(\frac{e^t-1}{t}\right) \log a= \log a$$

Answer 3

It is precisely $f'(0)$, where $f(r) = a^r$.

Answer 4

In order to evaluate the limit $$\lim_{x \to 0}\frac{a^{x} - 1}{x}\tag{1}$$ for $a > 0$, it is necessary to have a proper definition of $a^{x}$. This has been pointed out in one of the comments to the question, but OP has perhaps not understood the necessity of such a comment. In case you want to evaluate the limit $$\lim_{x \to 1}\frac{\sqrt{x} - 1}{x - 1}$$ it is necessary to have a definition of the symbol $\sqrt{x}$ and it is easy to define it by saying that $\sqrt{x}$ is a non-negative real number which gives $x$ when squared.

The unfortunate part here is that there is no definition of $a^{x}$ which is as simple as the definition of $\sqrt{x}$ and most introductory calculus texts try their best not to define this symbol in proper manner. The real sin is committed when they are able to convince the students that there is no necessity of such a definition. It is one thing to avoid a complicated definition but totally another matter to convince that such a thing does not exist.

There are multiple approaches to define $a^{x}$ and almost all of them are hard enough for a beginner in calculus (see these posts here in case you are interested).

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I prefer the following approach which is suitable for a beginner in calculus:

Theorem: For a given real number $a > 0$ there exists a unique function $F(x)$ defined and continuous for all real $x$ such that $F(0) = 1, F(1) = a$ and $F(x + y) = F(x)F(y)$ for all $x, y$. Moreover if $x$ is a rational number (say equal to $r \in \mathbb{Q}$) then $F(x) = F(r) = a^{r}$.

This function is called the general exponential function and denoted by symbol $a^{x}$. And then we come to the limit in question:

Theorem: For any real number $a > 0$ the limit $$\lim_{x \to 0}\frac{a^{x} - 1}{x}$$ exists and hence defines a function of $a$ (say $L(a)$) and this function $L(a)$ satisfies the following properties: $$L(1) = 0, L(xy) = L(x) + L(y)$$ Further $L(x)$ is continuous for all $x > 0$ and $$\lim_{x \to 0}\frac{L(1 + x)}{x} = 1\tag{2}$$ The function $L$ is traditionally denoted by $\log x$ and is called the natural logarithm (or just the logarithm) of $x$.

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Essentially the above approach gives two standard limits $$\lim_{x \to 0}\frac{a^{x} - 1}{x} = \log a,\,\lim_{x \to 0}\frac{\log(1 + x)}{x} = 1$$ and defers their proof (to advanced courses). But it is more honest than supplying vague ideas and intuition about exponential and logarithmic functions which I consider more of a hand waving.

Answer 5

Convert it into standard limit as write $a^x$ as $e^{x\ln a}$ and $\lim_{x \to 0} \dfrac{e^{x-1}}{x}=1$

Answer 6

Firt prove $\lim\limits_{x\to 0}\frac{e^x-1}x=1$. In fact, let $y=e^x-1$, that is $x=\ln(1+y)$, we have $$\lim\limits_{x\to 0}\frac{e^x-1}x=\lim\limits_{y\to 0}\frac{y}{\ln(1+y)} =\lim\limits_{y\to 0}\frac1{\frac1y\ln(1+y)}=\frac1{\ln\left[\lim\limits_{y\to 0} (1+y)^{1/y}\right]}=\frac1{\ln e}=1.$$ Then $$\lim\limits_{x\to 0}\frac{a^x-1}x=\lim\limits_{x\to 0}\frac{e^{x\ln a}-1}{x}= \ln a\lim\limits_{x\to 0}\frac{e^{x\ln a-1}}{x\ln a}=\ln a.$$