TITLE: Evaluate $\lim_{x \to \infty}{\frac{1}{2^x-5^x+3^x}}$. Why is my solution wrong?
QUESTION [5 upvotes]: Evaluate $$\lim_{x \to\infty}{\frac{1}{2^x-5^x+3^x}}. $$

My attempt:
$$\lim\limits_{x \rightarrow \infty}{\frac{1}{2^x-5^x+3^x}}
= \lim\limits_{x \rightarrow \infty}{\frac{1}{2^x} \cdot\frac{1}{1-\frac{5^x}{2^x}+\frac{3^x}{2^x}}} \\= \lim\limits_{x \rightarrow \infty}{\frac{1}{2^x} \cdot\frac{1}{\frac{1}{5^x}-\frac{5^x}{10^x}+\frac{3^x}{10^x}}\cdot\frac{1}{5^x}} = 0 
\cdot\infty \cdot 0 = 0.$$
Someone told me that it is a wrong way to show the limit. Can anyone explain why?

REPLY [4 votes]: $0\cdot \infty$ is $inderterminate$. To see this, calculate the following limits:
$$\lim_{x\to \infty} \frac{1}{x} \cdot x\qquad \text{and} \lim_{x\to \infty} 0\cdot x,$$
which are both of the form $0\cdot \infty$.