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

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