Thread: Limits Problem, can someone help with this?

1. Limits Problem, can someone help with this?

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?

2. Why can't you just substitute $x = \frac{\pi}{4}$?

You should know that $\tan{\frac{\pi}{4}} = 1$ and $\csc{\frac{\pi}{4}} = \frac{1}{\sin{\frac{\pi}{4}}} = \frac{1}{\frac{1}{\sqrt{2}}} = \sqrt{2}$.

3. Originally Posted by Prove It
Why can't you just substitute $x = \frac{\pi}{4}$?

You should know that $\tan{\frac{\pi}{4}} = 1$ and $\csc{\frac{\pi}{4}} = \frac{1}{\sin{\frac{\pi}{4}}} = \frac{1}{\frac{1}{\sqrt{2}}} = \sqrt{2}$.
Thank you, I don't know why I didn't think of that. I'm taking Calc for the first time as a summer class at my university and the professor is just flying by the course so I'm trying my best to keep up out of class

4. 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 $x= \frac{\pi}{4}$".

5. Here's another one I'm having trouble with

lim (s^3-y^3)/(s^4-y^4)
s->y

My professor didn't teach us the L'Hopital's rule, so I have absolutely no idea how to go about this one.

6. Why can't you just substitute s=y?

7. Originally Posted by Also sprach Zarathustra
Why can't you just substitute s=y?
This is actually a web assignment due and if I just substitue in y I get 0/0 which can't be done.

8. Try factoring?

$s^3 - y^3 = (s - y)(s^2 + sy + y^2)$
and
\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.
$\lim_{s \to y}\frac{s^3 - y^3}{s^4 - y^4}$
$= \lim_{s \to y}\frac{s^2 + sy + y^2}{(s + y)(s^2 + y^2)}$
$= \frac{y^2 + y^2 + y^2}{(2y)(y^2 + y^2)}$
$= \frac{3y^2}{4y^3}$
$= \frac{3}{4y}$

EDIT: Typing in LaTeX is #(\$*&#) difficult, grr...

9. Originally Posted by Nikhiln25
This is actually a web assignment due and if I just substitue in y I get 0/0 which can't be done.

Ohhh really?!

What you write first time was:

$lim_{s\to y}(s^3-y^3)/(s^4/y^4)$

Bravo!

10. Originally Posted by Also sprach Zarathustra
Ohhh really?!

What you write first time was:

$lim_{s\to y}(s^3-y^3)/(s^4/y^4)$

Bravo!
lol I realized and I edited sorry about that, I thought I was quick enough to update

11. Originally Posted by eumyang
Try factoring?

$s^3 - y^3 = (s - y)(s^2 + sy + y^2)$
and
\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.
$\frac{s^2 + sy + y^2}{(s + y)(s^2 + y^2)}$
My answer has to be in terms of y according to my assignment and if I substitute y in from here I end up getting y/2 as the limit which isn't showing up as right

12. Originally Posted by Nikhiln25
lol I realized and I edited sorry about that, I thought I was quick enough to update
^^ I was doing the same thing, editing while you were replying. Look at my post again.

13. Originally Posted by eumyang
^^ I was doing the same thing, editing while you were replying. Look at my post again.
Yeah that answer worked, and it makes a lot more sense now. I emailed my professor for help earlier and he told me to use synthetic division which got me lost even more. Thank you.

14. 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

15. ...

Here something useful...

$lim_{x\to 0} \frac{sin x}{x}=1$

$lim_{x\to 0} \frac{sin 4x}{9x}=\frac{4}{9}lim_{x\to 0} \frac{sin 4x}{4x}$