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lato1 · Dec 6, 2011, 10:24 PM

I have 3 very tricky limits questions

1 lim (x sin(x^2 e^x))/(x^2 1) as x approaches infinity.

2 lim(1/(e^x sin(1/x^2)) as x approaches infinity

3 lim(x-x^1/2 lnx) as x approaches infinity

Aurora2000 · Dec 7, 2011, 03:03 AM

You can try use de l'Hopital rule when dealing with limits of form $\frac{0}{0}, \frac{\infty}{\infty}$ :

1. $\lim_{x\rightarrow \infty} \frac{x\sin(x^2 e^x)}{x^2+1}$
Notice that $|\sin(x^2 e^x)|\leq 1$ , thus
$\lim_{x\rightarrow \infty} \frac{|x\sin(x^2 e^x)|}{x^2+1} \leq \lim_{x\rightarrow \infty} \frac{|x|}{x^2+1}=0$
thus
$\lim_{x\rightarrow \infty} \frac{x\sin(x^2 e^x)}{x^2+1}=0$

2. $\lim_{x\rightarrow \infty} \frac{1}{e^x\sin(1/x^2)}$
This is more tricky, you have to see who wins between $e^x$ (who goes to infinity) and
$\sin(1/x^2)$ (who goes to 0). Using
$\lim_{y\rightarrow 0}\frac{\sin y}{y}=1$
you can have
$\lim_{x\rightarrow \infty}\frac{\sin (1/x^2)}{1/x^2}=1$
thus
$\lim_{x\rightarrow \infty} \frac{1}{e^x\sin(1/x^2)}= \lim_{x\rightarrow \infty} \frac{1}{e^x\sin(1/x^2)}\frac{\sin(1/x^2)}{1/x^2}= \lim_{x\rightarrow \infty} \frac{x^2}{e^x}=0$

3. lim(x-x^1/2 lnx) as x approaches infinity
It does not display correctly (maybe a plus sign is missing), but both
$\lim_{x\rightarrow \infty} x-x^{1/2}+\log x$
and
$\lim_{x\rightarrow \infty} x-x^{1/2}\log x$
converge to infinity.
You can verify this by using
$\lim_{x\rightarrow \infty} x-x^{1/2}+\log x \geq \lim_{x\rightarrow \infty} \frac{x-x^{1/2}+\log x}{x^{3/4}}=\infty$
or similarly
$\lim_{x\rightarrow \infty} x-x^{1/2}\log x\geq \lim_{x\rightarrow \infty} \frac{x-x^{1/2}\log x}{x^{3/4}}=\infty$