1. ## Trig Limits

1.) limx-->0; (x + sinx) / (x + cosx)

2.) limx-->infinite; (e^-x)cosx

3. limx-->0; (xcscx + 1) / (xcscx)

thanks

2. ## not sure but

1.) limx-->0; (x + sinx) / (x + cosx)
dividing the top and bottom by x
=limx-->0; [1 + (sinx)/x] / [1 + (cosx)/x]
limx-->0 (sinx)/x=1 ;limx-->0 (cosx)/x=0
then continue... u should get 2

2.) limx-->infinite; (e^-x)cosx
=limx-->infinite; (cosx)/e^x
i think it is zero since limx-->infinite; e^x=infinite, but might also be undefined if x=n pi/2

3.) limx-->0; (xcscx + 1) / (xcscx)
divide top and bottom by xcsc+1
=limx-->0; [1+ (sinx)/x] / [(xcscx)/(xcscx)]
=limx-->0; [1+ (sinx)/x]
limx-->0 (sinx)/x=1; u should get 2

3. Hello, mushufasa11!

An alternate method for #3.

It uses the theorem: .$\displaystyle \lim_{\theta\to0}\frac{\sin\theta}{\theta} \:=\:1$

$\displaystyle 3)\;\;\lim_{x\to0}\frac{x\csc x + 1}{x\csc x}$

Multiply top and bottom by $\displaystyle \sin x$

Then we have: .$\displaystyle \lim_{x\to0}\frac{x + \sin x}{x} \;=\;\lim_{x\to0}\left(1 + \frac{\sin x}{x}\right) \;=\;1 + 1 \;=\;2$