I’m trying to solve the limit:

limx→∞x3+2xx2+3x

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**

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.

**Attribution***Source : Link , Question Author : calcstudent , Answer Author : StackTD*