Seems Familiar

Calculus Level 1

$\large \lim_{n\to\infty} \left(1 - \frac{x^2}{n^2} \right)^n = \ ?$

The answer is 1.

1 solution

Curtis Clement
May 9, 2015

$\lim_{n \rightarrow\infty } (1- \frac{x^2}{n^2})^n = \lim_{n \rightarrow\infty } (1+\frac{x}{n})^n \times\lim_{n \rightarrow\infty } (1- \frac{x}{n})^n = e^x \times\ e^{-x} = \boxed{1}$

This comes from the definition of e: $e = \lim_{n \rightarrow\infty } (1+\frac{1}{n})^n$ which can be generalised to $\ e^x = \lim_{n \rightarrow\infty } (1+\frac{x}{n})^n \ ........ x \in \Re$ via l'hopital's rule.

I'm not really into limits yet I was able to answer this question correctly(by mistake I suppose) but I wanted to know where I've gone wrong.

So IMO there's no need for all that explanation when $n \rightarrow \infty$ then the denominator in $\frac {x^2}{n^2}$ tends to infinity which means the denominator is zero in the infinity.Hence the answer should be $\lim_{n \rightarrow\infty } (1^n)$ which is 1.By the way I don't really know much about the $e$ so if the wrong in my answer was anything somehow related to $e$ please explain it noob-friendly xD thanks for the problem.

Arian Tashakkor - 6 years, 1 month ago

That would usually be the case, but the trouble is the exponent is getting larger at the same time, so we to look for convergence. There is a neat proof that $\lim_{n \rightarrow\infty} (1+\frac{1}{n})^n$ monotonically converges to e $\approx$ 2.718. If you are still not convinced just type in the expression for e as n gets larger.

Curtis Clement - 6 years, 1 month ago

Thank you very much Curtis!I think I get it now.And also the reason for the name "Seems familiar ... " :D

Arian Tashakkor - 6 years, 1 month ago

Seems Familiar

The answer is 1.

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