Determine a limit of a product Suppose $f(x)\to 0$ as $x\to\infty$.
I would like to show that
$$
\lim_{x\to\infty}f(x)(1+f(x))^{2x}=0
$$
Is this trivial because
$$
\lim_{x\to\infty}f(x)(1+f(x))^{2x}=\lim_{x\to\infty}f(x)\cdot\lim_{x\to\infty}((1+f(x))^{2x})
$$
and
$$
\lim_{x\to\infty}((1+f(x))^{2x})=\underbrace{\overbrace{\lim_{x\to\infty}(1+f(x))}^{=1}\cdot\ldots\cdot\overbrace{\lim_{x\to\infty}(1+f(x))}^{=1}}_{2x-times}=1
$$
I think this is not valid since as $x\to\infty$, the number of factors goes to infinity.
 A: Let $f(x)=x^{-1/2}$.
Then
$$\lim_{x\to\infty}x^{-1/2}(1+x^{-1/2})^{2x}=\lim_{t\to\infty}t^{-1}(1+t^{-1})^{2t^2}=\lim_{t\to\infty}t^{-1}((1+t^{-1})^t)^{2t}.$$
The last expression is bounded below by $\dfrac{2^{2t}}t$ (for $t\ge1$) and the limit diverges.

Note that with $f(x)=x^{-1}$ you bound above with $\dfrac9x$ and the limit is indeed $0$. It is much more challenging (though IMO possible) to find a function that leads to a finite limit.
A: There are two problems in your interpretation of $(1+f(x))^{2x}$.

*

*When $x$ is a real number, the expression $a^x$ ($a>0, a\ne 1$) is not interpreted as the multiplication of "$x$ many $a$'s". It does not make sense to say multiply $\pi$ copies of $a$. Instead, one of its definitions is
$$
a^x=e^{x\ln a}
$$

*Even you restrict $x$ to be positive integers,
$$
\lim_{n\to\infty}(a_n)^{n}\ne \lim_{n\to\infty}(\lim_{n\to\infty}a_n)^{n}.
$$

The statement you want to show:

if $\lim_{x\to\infty}f(x)=0$, then $\lim_{x\to\infty}f(x)(1+f(x))^{2x}=0$

is in general not true. The problem is that one of the factor is of the indeterminant form: $1^\infty$. One counterexample has been given in a comment and Yves's answer.
As a convenient sufficient condition, if the function $f$ is in addition such that
$$
\lim_{x\to\infty}x\ln((1+f(x))=0
$$
then by continuity, you have
$$
\lim_{x\to\infty}\exp\big(2x\ln((1+f(x))\big)=1
$$
