How to solve $\lim_{x \rightarrow 0} {\frac{(1+x)^a-1}{x}}$?

Question

How to solve this limit without using L'Hospital rule?

$$\lim_{x \rightarrow 0}{\frac{(1+x)^a-1}{x}}$$

Answer 1

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

Answer 2

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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.

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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

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

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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!