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?

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- Jul 5th 2010, 08:21 PMNikhiln25Limits 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? - Jul 5th 2010, 08:24 PMProve It
Why can't you just substitute $\displaystyle x = \frac{\pi}{4}$?

You should know that $\displaystyle \tan{\frac{\pi}{4}} = 1$ and $\displaystyle \csc{\frac{\pi}{4}} = \frac{1}{\sin{\frac{\pi}{4}}} = \frac{1}{\frac{1}{\sqrt{2}}} = \sqrt{2}$. - Jul 5th 2010, 08:30 PMNikhiln25
- Jul 6th 2010, 01:19 AMHallsofIvy
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 $\displaystyle x= \frac{\pi}{4}$".

- Jul 6th 2010, 12:07 PMNikhiln25
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. - Jul 6th 2010, 12:13 PMAlso sprach Zarathustra
Why can't you just substitute s=y?

- Jul 6th 2010, 12:19 PMNikhiln25
- Jul 6th 2010, 12:22 PMeumyang
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... - Jul 6th 2010, 12:27 PMAlso sprach Zarathustra
- Jul 6th 2010, 12:29 PMNikhiln25
- Jul 6th 2010, 12:35 PMNikhiln25
- Jul 6th 2010, 12:37 PMeumyang
- Jul 6th 2010, 12:42 PMNikhiln25
- Jul 7th 2010, 06:21 PMNikhiln25
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 - Jul 7th 2010, 06:47 PMAlso sprach Zarathustra...
Here something useful...

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

in your case (q. a):

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

Try apply this also for b.