# Proving (1+x)^n approaches 1 + nx when x goes to zero

1. May 13, 2012

### U.Renko

1. The problem statement, all variables and given/known data
I've heard this affirmation in an gravitation lecture. I thought it was interesting and decided to check it out. It turns out to be true so I decided to prove it.

I'm not a mathematician so proving stuff is not my department. Still I try to do some proofs just for fun. I hope I don't make your eyes bleed or something.

2. Relevant equations
$\lim_{x \rightarrow 0} (1 + x)^n$

3. The attempt at a solution
I'm not sure if I should try prooving by epsilon-delta or by mathematical induction (if that is even possible)
I did try by mathematical induction but it didnt look convincing

How would you do it?

2. May 13, 2012

### Infinitum

$\lim_{x \to 0} (1 + x)^n$

Apply binomial expansion to this equation and then the concept of limits.

3. May 13, 2012

### Curious3141

Try Binomial Theorem.

While it is correct to say ${(1+x)}^n \approx 1+nx$ for sufficiently small x, the actual limit of the expression as x tends to zero is simply 1.

4. May 13, 2012

### micromass

It's impossible to prove this until you rigorized what the statement means.

I think the best rigorous version of this statement is that y=1+nx is the tangent line of y=(1+x)n at the point x=0.

Or, it could mean

$$\lim_{x\rightarrow 0} \frac{(1+x)^n-(1+nx)}{x}=0$$

which means that the functions are equal up to first order.

Try to prove these two statements.