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

[U.Renko]

Homework Statement

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.

Homework Equations

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

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?

 

[Infinitum]

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

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

 

[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.

 

[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.