Lim
x-> Pi/4
1+tan(x)
---------
csc(x)+2
I'm guessing you can take the lim of 1, lim of tan(x) and divide it by the lim of csc(x) + 2?
As long as the denominator is not 0, taking the limit of each part separately is just using the various "laws of limits" and is valid. If each part is a continuous function (and the denominator is not 0 so the whole combination is a continous function) then that is exactly the same as "substitute $\displaystyle x= \frac{\pi}{4}$".
Try factoring?
$\displaystyle s^3 - y^3 = (s - y)(s^2 + sy + y^2)$
and
$\displaystyle \begin{aligned}
s^4 - y^4 &= (s^2 - y^2)(s^2 + y^2) \\
&= (s - y)(s + y)(s^2 + y^2)
\end{aligned}$
Then the (s - y) factors cancel out.
$\displaystyle \lim_{s \to y}\frac{s^3 - y^3}{s^4 - y^4}$
$\displaystyle = \lim_{s \to y}\frac{s^2 + sy + y^2}{(s + y)(s^2 + y^2)}$
$\displaystyle = \frac{y^2 + y^2 + y^2}{(2y)(y^2 + y^2)}$
$\displaystyle = \frac{3y^2}{4y^3}$
$\displaystyle = \frac{3}{4y}$
EDIT: Typing in LaTeX is #($*&#) difficult, grr...
Can someone tell me how to solve these kind of problems? I have a couple of these in my homework, and had one on a quiz today but couldn't really figure out how to solve them
1)
lim sin4x/9x
x->0
2)
lim
x->0 tan4x/2x
I tried to put it into wolfram alpha but that used the L'Hopital's rule which made no sense to me since the professor didn't teach L'Hopital's rule to my class