2

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.

Tim Raczkowski
  • 20,046
user253724
  • 103

4 Answers4

3

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 $0$.

preferred_anon
  • 17,121
0

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 $x\to -\infty$.)

Joel
  • 16,256
  • When When x→+∞ then −ln(3)x2→−∞ and ln(2)x→+∞ which leads in an uknown form of ( -∞ +∞) ... – user253724 Jul 15 '15 at 18:54
  • Don't think of a polynomial as its individual terms. The only term that matters for large $x$ is the leading one. Thus $x^3 - x^2 \to \infty$ as $x\to \infty$ and $-x^2+x \to -\infty$ as $x\to\infty$. @user253724 – Joel Jul 15 '15 at 20:27
  • In this specific case we can actually graph $f(x) = -\ln(3)x^2 + \ln(2) x = x(\ln(2) - \ln(3)x)$. This is a parabola facing downwards, with zeros $x=0$ and $x = \ln(2)/\ln(3)$. – Joel Jul 15 '15 at 20:29
0

Hint: When $x\to+\infty$, note $$\frac{2^x}{3^{x^2} } = \left(\frac{2}{3^x}\right)^x.$$ When $x\to -\infty$, note $$ \frac{2^x}{3^{x^2} } = 2^x 3^{- x^2}.$$

Vincenzo Oliva
  • 7,249
0

$$\lim\limits_{x\to\infty} \frac{2^x}{3^{x^2}}$$ 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 $$\lim\limits_{x\to\infty} \frac{2^x}{3^{x^2}}=0$$

k170
  • 9,045