How to solve $\lim_{x \rightarrow 0} {\frac{(1+x)^a-1}{x}}$? How to solve this limit without using L'Hospital rule?
$$\lim_{x \rightarrow 0}{\frac{(1+x)^a-1}{x}}$$
 A: Using the generalized binomial expansion, we get
$$\lim_{x \rightarrow 0}{\frac{-1 + \sum_{k=0}^\infty {a\choose k}x^k }{x}}$$
$$=\lim_{x \rightarrow 0}{\frac{-1 + 1 + ax + {a\choose 2}x^2 + {a\choose 3}x^3 +\cdots }{x}}$$
$$=a + \lim_{x \rightarrow 0}{\frac{{a\choose 2}x^2 + {a\choose 3}x^2 +\cdots }{x}}$$
$$=a + \lim_{x \rightarrow 0}\left[{a\choose 2}x + {a\choose 3}x^2 +\cdots\right]$$
$$=a$$  
Note the dependence of $x$ on the entire RHS of the limit in the second-to-last equation
A: 

PRIMER:
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
$$1+x\le e^x\le \frac{1}{1-x} \tag 1$$
for $x<1$, and
$$\frac{x-1}{x}\le \log(x)\le x-1\tag 2$$
for $x>0$.  We will use the inequalities in $(1)$ and $(2)$ to evaluate the limit of interest.


First note that using $(2)$, it is easy to see that for $x>0$ we have
$$\frac{1}{1+x} \le \frac{\log(1+x)}{x}\le 1 \tag 3$$
while for $-1<x<0$ we have
$$1 \le \frac{\log(1+x)}{x}\le \frac{1}{1+x} \tag 4$$
Applying the squeeze theorem to $(3)$ and $(4)$, we see that
$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0}\frac{\log(1+x)}{x}=1} \tag 5$$

Next, we write $(1+x)^a=e^{a\log(1+x)}$ so that
$$\frac{(1+x)^a-1}{x}=\frac{e^{a\log(1+x)}-1}{x}$$
Now, we restrict $x$ such that $a\log(1+x)<1$.  Then, we exploit $(1)$ to reveal
$$\frac{a\log(1+x)}{x}\le \frac{(1+x)^a-1}{x}\le \frac{a\log(1+x)}{x(1-a\log(1+x))} \tag 6$$
Using $(5)$ and applying the squeeze theorem to $(6)$, we obtain
$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0}\frac{(1+x)^a-1}{x}=a}$$
And we are done!  No series, no L'Hospital's, and no derivatives!
