# Evaluate $\lim_{x \to \infty}{\frac{1}{2^x-5^x+3^x}}$. Why is my solution wrong?

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?

• Because $0\cdot\infty$ is an indeterminate form...? – CiaPan Aug 18 '17 at 7:47
• You have problem when you go from limit expression to $0\cdot\infty\cdot 0$. This is not allowed via any theorem on limits. – Paramanand Singh Aug 18 '17 at 8:12

Because $0\cdot \infty$ is an Indeterminate form. Note that $${\frac{1}{2^x-5^x+3^x}}=\left(\frac{1}{5}\right)^x\cdot \frac{1}{\left(\frac{2}{5}\right)^x-1+\left(\frac{3}{5}\right)^x}.$$ Since $1/5$, $2/5$, and $3/5$ are positive numbers less than $1$, what may we conclude as $x\to +\infty$?
• Obviously 0*(-1) = 0. I did not know about indetermined form yet neither I had the idea of pull the $5^x$ out of the fracture. Thanks for fixing my issue. – Maddude Aug 18 '17 at 7:59
$0\cdot \infty$ 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 $0\cdot \infty$.
Because it is not true that $0\times\infty\times0=0$. In fact, $0\times\infty\times0$ is indeterminate.
• Actually "$0\times\infty\times0$" is indeterminate, not "$0\times\infty\times0=0$". – CiaPan Aug 18 '17 at 7:51