Limit of a function involving a sequence.

Question

I have the following problem:

Suppose that $\lim_{n \to \infty} a_n = 0$. Prove that for any $x$ $$\lim_{n \to \infty} \left(1+a_n \frac{x}{n}\right)^n = 1.$$

I have tried replacing $a_n$ with a function $a(n)$ and applying L'Hopital's Rule but got no useful result. I also tried going directly to the definitions and showed that $\lim_{n \to \infty} (1+a_n \frac{x}{n}) = 1$, but I don't know how to deal with the $n$th power (that makes the limit $\lim_{n \to \infty} \exp(n(1+a_n \frac{x}{n}))$ indeterminate).

Answer 1

Fix $x\in\mathbf R$.

By Bernoulli's inequality (here $n$ has to be large enough, so that $a_n x/n>-1$), $$ \Bigl(1+\frac{a_nx}{n}\Bigr)^n\geq 1+a_n x. $$ On the other hand, since $$ \Bigl(1+\frac{a}{n}\Bigr)^n\leq e^a $$ for all $a\in\mathbf R$, $$ \Bigl(1+\frac{a_nx}{n}\Bigr)^n\leq e^{a_n x}. $$ Now use the sandwich theorem.

Answer 2

Take log: $$ \lim_{n\to\infty} n\ln\left(1 + a_n\frac{x}{n}\right). $$ But if $a_n\to0$ $$ \ln\left(1 + a_n\frac{x}{n}\right)\sim a_n\frac xn, $$ and $$ \lim_{n\to\infty} n\ln\left(1 + a_n\frac{x}{n}\right) = \lim_{n\to\infty} a_n x = x\lim_{n\to\infty} a_n = 0; $$ so $$ \lim_{n\to\infty} \left(1 + a_n\frac{x}{n}\right)^n = 1 $$

Answer 3

Notice, $$\lim_{n\to \infty}\left(1+a_n\frac{x}{n}\right)^n$$ Using binomial expansion of $\left(1+a_n\frac{x}{n}\right)^n$ & neglecting higher power terms, we get $$=\lim_{n\to \infty}\left(1+na_n\frac{x}{n}\right)$$ $$=\lim_{n\to \infty}\left(1+a_n x\right)$$ $$=\lim_{n\to \infty} 1+x\lim_{n\to \infty}a_n $$ $$= 1+x(0)=1 $$

Answer 4

I am giving an simple answer.

$\lim\limits_{x\to \infty} a_n \to 0$

Now, $\lim\limits_{x \to \infty} (1+a_n \frac xn)^n =\lim\limits_{x\to \infty}((1+a_n \frac xn)^ {\frac n{a_n x}})^{a_nx} =e^{\lim\limits_{n\to \infty}\;a_n x} =e^0=1$

Answer 5

$$ \left(1 + \frac{a_n x}{n}\right)^n = \bigg[ \underbrace{\left(1+\frac{a_n x}{n}\right)^{\frac{n}{a_n x}}}_{\to e} \bigg]^{\overbrace{a_n x}^{\to 0}} \to e^0 = 1 $$