Solving limit with unknown exponent without L'Hopital

Question

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., Apr 21 '17 at 21:57
Are you not using binomial theorem because you don't know whether $a \in \mathbb N $ or not. Right? — Jaideep Khare, Apr 21 '17 at 22:01
Then let me edit my answer.You can use extended binomial theorem dor any $a \in \mathbb R$ — Jaideep Khare, Apr 21 '17 at 22:03
@bthmas not in this context.. Sorry guys I'm pretty new to calculus. — Caio Petrelli, Apr 21 '17 at 22:05
If you are new to calculus, it is best to prove the result for rational $a$ using the techniques of algebraic manipulation or if you are adventurous enough you can take inequalities / Squeeze Theorem approach also. The case for irrational $a$ has to be deferred till the time when you have understood the theory of exponential and logarithmic functions (see Dr. MV's answer which uses this). — Paramanand Singh, Apr 22 '17 at 03:35
Also note that this limit is in form of derivative so using L'Hospital's Rule here is kind of very roundabout. — Paramanand Singh, Apr 22 '17 at 03:37

score 5 · Accepted Answer · edited Jun 12 '20 at 10:38

5

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

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

edited Jun 12 '20 at 10:38

Community

1

answered Apr 21 '17 at 22:07

Mark Viola

179,405

Oh, I see it now! Thanks! – Caio Petrelli Apr 21 '17 at 22:09
You're welcome! My pleasure and pleased I could help you. -Mark – Mark Viola Apr 21 '17 at 22:10
+1 This was the approach I was alluding to in my comment. – bthmas Apr 21 '17 at 22:11
@bthmas Thank you; much appreciative! -Mark – Mark Viola Apr 21 '17 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 – Caio Petrelli Apr 21 '17 at 22:16
Caio, that's great! I'm glad that this is fit for purpose. – Mark Viola Apr 21 '17 at 22:17

Jaideep Khare · Answer 2 · 2017-04-21T22:19:00.890

1

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

edited Apr 21 '17 at 22:19

answered Apr 21 '17 at 21:57

Jaideep Khare

19,293

2

OP says he cannot use the Binomial th. – caverac Apr 21 '17 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 – Jaideep Khare Apr 21 '17 at 22:00
It's much easier to read this quotient as a rate of variation. – Bernard Apr 21 '17 at 22:04
@CaioPetrelli See this.This can be used for any $a$. – Jaideep Khare Apr 21 '17 at 22:07
@JaideepKhare I can't really see how that makes it easier.. – Caio Petrelli Apr 21 '17 at 22:11
@CaioPetrelli See now. – Jaideep Khare Apr 21 '17 at 22:19
@JaideepKhare Oh, I see it. That's a very beautiful approach.. I would never seen that, thanks! – Caio Petrelli Apr 21 '17 at 22:23
The binomial theorem for any index is proved using that limit in question. So this is kind of circular. – Paramanand Singh Apr 22 '17 at 03:07

score 1 · Answer 3 · answered Apr 21 '17 at 22:05

1

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.

answered Apr 21 '17 at 22:05

Patrick Stevens

36,135

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! – Caio Petrelli Apr 21 '17 at 22:15

score 1 · Answer 4 · answered May 01 '17 at 06:38

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. – Paramanand Singh May 01 '17 at 07:16 — Paramanand Singh, May 01 '17 at 07:16

score 0 · Answer 5 · answered Apr 22 '17 at 15:28

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

Solving limit with unknown exponent without L'Hopital

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