Simple limit problem with $a^x$. Any idea of calculating the limit of $f(x)=2^x/3^{x^2}$ when $x$ approaches infinate? I looked up the function in geogebra and the limit is zero. Not sure how to prove it though... I guess using the L' Hospital (not sure if spelled correctly) rule.
 A: For $x>1$, $x^{2}>x$, so $$0<\frac{2^{x}}{3^{x^{2}}}<\frac{2^{x}}{3^{x}} \to 0$$
So the sequence has to tend to $0$.
A: Notice that $2^x = e^{\ln(2)x}$ and $3^{x^2} = e^{\ln(3)x^2}$. Thus $$\frac{2^{x}}{3^{x^2}} = e^{-\ln(3)x^2 + \ln(2) x}.$$
Thus as $x\to \infty$ we have $-\ln(3)x^2+\ln(2)x \to -\infty$ (since $\ln(3) > 0$) so we conclude $$\frac{2^x}{3^{x^2}} =  e^{-\ln(3)x^2 + \ln(2) x}\to 0$$ as $x\to \infty$. (Remember $e^{x} \to 0$ as $x\to -\infty$.)
A: Hint: When $x\to+\infty$, note $$\frac{2^x}{3^{x^2} } = \left(\frac{2}{3^x}\right)^x.$$
When $x\to -\infty$, note $$ \frac{2^x}{3^{x^2} } = 2^x 3^{- x^2}.$$ 
A: $$\lim\limits_{x\to\infty} \frac{2^x}{3^{x^2}}$$
For $x\geq 1$, we have
$$2\leq 2^x\leq 2^{x^2}$$
$$\frac{2}{3^{x^2}}\leq\frac{2^x}{3^{x^2}} \leq \frac{2^{x^2}}{3^{x^2}}$$
$$\lim\limits_{x\to\infty}\frac{2}{3^{x^2}}\leq\lim\limits_{x\to\infty}\frac{2^x}{3^{x^2}} \leq \lim\limits_{x\to\infty}\frac{2^{x^2}}{3^{x^2}}$$
$$0\leq\lim\limits_{x\to\infty}\frac{2^x}{3^{x^2}} \leq 0$$
Therefore 
$$\lim\limits_{x\to\infty} \frac{2^x}{3^{x^2}}=0$$
