Solving limit with unknown exponent without L'Hopital I came across this limit and it was told that we should not use L'Hopital's rules for this one:
$$
\lim _{x\to 0\:}\left(\frac{\left(1+x\right)^a-1}{x}\right)
$$
I can't see a way to get around that x on the denominator, I can't expand the binomial because it's an unknown.. Any solutions? Is it even possible without L'Hopital?
 A: We will evaluate the limit using only standard inequalities and the squeeze theorem.  To that end, we begin with a short primer.


PRIMER ON STANDARD INEQUAITIES
In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm and exponential functions satisfy the inequalities
$$\frac{x-1}{x}\le\log(x)\le x-1 \tag 1$$
for $x>0$ and
$$1+x\le e^x\le \frac{1}{1-x}\tag 2$$
for $x<1$


First, note that we can write $(1+x)^a=e^{a\log(1+x)}$.  Then, using $(1)$ and $(2)$ we find that for $ax<1$
$$\frac{a}{x+1}\le\frac{(1+x)^a-1}{x}\le \frac{a}{1-ax}\tag3$$
whence applying the squeeze theorem to $(3)$ yields the coveted limit

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0}\frac{(1+x)^a-1}{x}=a}$$

A: Use binomial theorem for any index.
$$(1+x)^a=1+ax+\frac{a(a-1)}{2!}x^2+\frac{a(a-1)(a-2)}{3!}x^3+\frac{a(a-1)(a-2)(a-3)}{4!}x^4 .....$$
$$\implies \lim _{x\to 0\:}\left(\frac{\left(1+x\right)^a-1}{x}\right)=\lim _{x\to 0\:}\left(\frac{\big(1+ax+\frac{a(a-1)}{2!}x^2+ \ldots)-1}{x}\right)$$
$$=\lim _{x\to 0\:}\left(a+\frac{a(a-1)}{2!}x+\ldots \right)=\boxed a$$
$\big($ Here $|x| <1$ and $a \in \mathbb R$  $\big)$
A: Whenever there's $h \to 0$ on the denominator, you should think of the derivative. You have written the derivative of $y \mapsto (1+y)^a$ at $y=0$, which you can easily calculate by the chain rule.
A: Let $z=x+1$, then $x\to 0$ as $z\to 1$, and for change of variable:
$$\lim_{x\to0}\frac{(1+x)^{a}-1}{x}=\lim_{z\to1}\frac{z^{a}-1}{z-1}$$
Using the fact that $a^{n}-b^{n}=(a-b)\left(\displaystyle\sum_{i=0}^{n-1}a^{n-1-i}b^{i}\right)$, therefore:
$$\lim_{z\to1}\frac{z^{a}-1}{z-1}=\lim_{z\to1}\frac{(z-1)\left(\displaystyle\sum_{i=0}^{a-1}z^{a-1-i}\right)}{z-1}=\lim_{z\to1}\left(\displaystyle\sum_{i=0}^{a-1}z^{a-1-i}\right)=\displaystyle\sum_{i=0}^{a-1}\lim_{z\to1}z^{a-1-i}=\sum_{i=0}^{a-1}1=a$$
As desidered.
A: Just in case you want to see another approach, using
$$\lim_{x\to0}\frac{e^x-1}{x} = 1$$ and $$\lim_{x\to0}\frac{\ln(1+x)}{x}=1$$
Therefore, $$\lim_{x\to0}\frac{(1+x)^a-1}{x} = \lim_{x\to0}\frac{e^{\ln(1+x)^a}-1}{\ln(1+x)^a}\cdot\frac{a\ln(1+x)}{x} = a$$
