Why does this limit equal 0?

Question

I'm trying to solve the limit: $$\lim_{x \to \infty}\frac{x^3+2^x}{x^2+3^x}$$ So far I've divided every term by $3^x$ and this is correct according to the solution, which gets me: $$\lim_{x \to \infty}\frac{\frac{x^3}{3^x}+( \frac{2}{3})^x}{\frac{x^2}{3^x}+1}$$ I would've said that the $X^3/3^x$ goes to 0 and the $(2/3)^x$ goes to infinity and then $x^2/3^x$ goes to 0 and the one stays as it is. This would give $\frac{0+\infty}{0+1}$ which is the same as $\frac{\infty}{1}$ which is infinity.

However, the answer according to the solution is 0; could anyone explain why this is?

Thanks

Answer 1

Your error lies here:

I would've said that the $X^3/3^x$ goes to 0 and the $(2/3)^x$ goes to infinity

The asymptotic behaviour of the exponential function $a^x$ depends on the base $a$:

• if $a>1$, then $a^x$ is a strictly increasing function and $a^x \to +\infty$ for $x \to +\infty$;
• if $0<a<1$, then $a^x$ is a strictly decreasing function and $a^x \to 0$ for $x \to +\infty$.

In your case $\left(\tfrac{2}{3}\right)^x \to 0$ for $x \to +\infty$ because $0<\tfrac{2}{3}<1$.