# 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?

• Use the derivative definition $\lim_{x\to 0}\frac{f(1+x)-f(1)}{x}$
– A.Γ.
Commented Apr 21, 2017 at 21:57
• Are you not using binomial theorem because you don't know whether $a \in \mathbb N$ or not. Right? Commented Apr 21, 2017 at 22:01
• @JaideepKhare exactly Commented Apr 21, 2017 at 22:02
• Have you heard of squeeze theorem? Commented Apr 21, 2017 at 22:02
• Then let me edit my answer.You can use extended binomial theorem dor any $a \in \mathbb R$ Commented Apr 21, 2017 at 22:03

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

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

• Oh, I see it now! Thanks! Commented Apr 21, 2017 at 22:09
• You're welcome! My pleasure and pleased I could help you. -Mark Commented Apr 21, 2017 at 22:10
• +1 This was the approach I was alluding to in my comment. Commented Apr 21, 2017 at 22:11
• @bthmas Thank you; much appreciative! -Mark Commented Apr 21, 2017 at 22:13
• I think this is a very suiting answer because that uses things that i've seen in the course thanks again @Dr.MV Commented Apr 21, 2017 at 22:16

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.

• I see where you're coming from but unfortunately I'm still looking at limits as being just limits. I'm still early in the course! But thanks that made me appreciate the simplicity of things! Commented Apr 21, 2017 at 22:15

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 $a \in \mathbb R$ $\big)$

• OP says he cannot use the Binomial th. Commented Apr 21, 2017 at 21:58
• @caverac He said because he is not sure whether $a$ is an natural number, I am saying to use for any index Commented Apr 21, 2017 at 22:00
• It's much easier to read this quotient as a rate of variation. Commented Apr 21, 2017 at 22:04
• @CaioPetrelli See this.This can be used for any $a$. Commented Apr 21, 2017 at 22:07
• @JaideepKhare I can't really see how that makes it easier.. Commented Apr 21, 2017 at 22:11

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.

• $a$ is not necessarily a positive integer. Commented May 1, 2017 at 7:16

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, $$\lim_{x\to0}\frac{(1+x)^a-1}{x} = \lim_{x\to0}\frac{e^{\ln(1+x)^a}-1}{\ln(1+x)^a}\cdot\frac{a\ln(1+x)}{x} = a$$