okay.
I am really bad at limits. I'll state that up front.
d) I agree that d is undefined thaough I do not know what d.n.e stands for. 
f) I get the same as Chris too.
$$\displaystyle\lim_{x\rightarrow\infty}\;\;\frac{x+1}{2x^2+1}\\\\
=\displaystyle\lim_{x\rightarrow\infty}\;\;\frac{1}{4x+1}\\\\
=0\\\\\\
=\displaystyle\lim_{x\rightarrow\infty}\;\;e^0\\\\
=1$$
e) I'll admit to learning this one straight from CPhill. Thanks Chris
$$\displaystyle\lim_{x\rightarrow\infty}\;\frac{3-cosx}{x^2+10}}\\\\
=\displaystyle\lim_{x\rightarrow\infty}\;\frac{3}{x^2+10}}\;\;-\;\;\displaystyle\lim_{x\rightarrow\infty}\;\frac{cosx}{x^2+10}}\\\\
=\;0\;-\;\;\displaystyle\lim_{x\rightarrow\infty}\;\frac{cosx}{x^2+10}}\\\\
=\;-\;\;\displaystyle\lim_{x\rightarrow\infty}\;\frac{cosx}{x^2+10}}\\\\
$Now $-1\le cosx\le 1\\\\
$So $\\\\
\displaystyle\lim_{x\rightarrow\infty}\;\frac{-1}{x^2+10}\;\;\le \displaystyle\lim_{x\rightarrow\infty}\;\frac{cosx}{x^2+10}\;\le\displaystyle\lim_{x\rightarrow\infty}\;\frac{1}{x^2+10}\\\\
0\;\;\le \displaystyle\lim_{x\rightarrow\infty}\;\frac{cosx}{x^2+10}\;\le\;0\\\\
$therefore$\\\\
\displaystyle\lim_{x\rightarrow\infty}\;\frac{cosx}{x^2+10}\;=\;0\\\\$$
$$\\$therefore$\\\\
\displaystyle\lim_{x\rightarrow\infty}\;\frac{3-cosx}{x^2+10}}=0-0=0\\\\$$
Thank you for showing me that Chris 
(This answer is exactly the same as Chris's)
+118733 
