Question

How do I solve for this limit?

Answer 1

\[\lim_{x \rightarrow \infty}\frac{ 2^x }{ 3^x-2^x }\]

Answer 2

HINT: Divide numerator and denominator by \(3^x\).

Answer 3

So you then would have \(\displaystyle\lim_{x\to\infty}\dfrac{\left(\dfrac{2}{3}\right)^x}{\dfrac{3^x-2^x}{3^x}} = \lim_{x\to\infty}\dfrac{\left(\dfrac{2}{3}\right)^x}{1-\left(\dfrac{2}{3}\right)^x} = \dfrac{0}{1-0} = \boxed{0}\)

Since we know that \(\displaystyle \lim_{x\to\infty}a^x = 0\) for \(-1<a<1\)

Answer 4

ditto

Answer 5

Yes?

Answer 6

Thanks : )