Find $\lim\limits_{x \to \infty}\frac{2^x}{3^{x^2}}$ I can only reason with this intuitively. since $3^{x^2}$ grows much faster than $2^x$ the limit as $x \to \infty$ of $f$ must be 0.
Is there a more rigorous way to show this?
Find $\lim\limits_{x \to \infty}\frac{2^x}{3^{x^2}}$ I can only reason with this intuitively. since $3^{x^2}$ grows much faster than $2^x$ the limit as $x \to \infty$ of $f$ must be 0.
Is there a more rigorous way to show this?
After so many equivalent examples one more should not be missed because of its form. Here we get the numerator constant and keep the notation in powers of 2 and 3. Denote ß=\frac{\ln(2)}{\ln(3)} Then \frac{2^x}{3^{x^2}}=3^{ßx-x^2}=3^{(ß/2)²-(x-\fracß2)^2} = \frac{2^{ß/4}}{3^{(x-ß/2)^2}} where the numerator is constant. We see also, that the denominator is minimal if x=ß/2 and the whole expression is maximal by the exponent ßx-x^2 of the second term, its first derivative ß-2x (which is zero at $x=ß/2$) and its second derivative $-2$ which is negative and shows, that the whole expression has a maximum there.
How about comparison to $\displaystyle{\frac{2^x}{3^x}=\left(\frac{2}{3}\right)^x}$?
Hint: $2^x \lt 3^x$ for sufficiently large $x$.
How about rewriting $\frac{2^x}{3^{x^2}}$ as $\frac{2^x}{3^x3^{x^2-x}}=(\frac{2}{3})^x\cdot\frac{1}{3^{x^2-x}}$?
Rewriting the fraction as $\frac{e^{x \ln 2}}{e^{x^2 \ln 3}} = e^{x \ln 2 - x^2 \ln 3}$ may also help you (although perhaps no more than the other answers given already).