If : $l_1 = \lim_{x \rightarrow 2} \left( x + |x| \right) $

$l_2 = \lim_{x\rightarrow 2} (2x - |x| ) $

and

$l_3 = \lim_{x \rightarrow \frac{\pi}{2} } \left( \cfrac{ \cos x}{x - \cfrac{\pi}{2}} \right) $

Find the relationship between $l_1$ , $l_2$ and $l_3$ .

Question

If : $l_1 = \lim_{x \rightarrow 2} \left( x + |x| \right) $ 

$l_2 = \lim_{x\rightarrow 2} (2x - |x| ) $ 

and 

$l_3 = \lim_{x \rightarrow \frac{\pi}{2} } \left( \cfrac{ \cos x}{x - \cfrac{\pi}{2}} \right) $ 

Find the relationship between $l_1$ , $l_2$ and $l_3$ .

mathslover · Answer

It might not be tough for @hartnn , @mukushla or @Miracrown 
but, this is the question, where I'm stuck.

Its not a challenge now ..!

hartnn · Answer

whats stopping you from direct substitution in l1 and l2 ?

miracrown · Answer

i take it as a challenge :P

anonymous · Answer

:-p

miracrown · Answer

Well, I suppose if we just compute them, we can come up with any number of relationships we want.

mathslover · Answer

@hartnn didn't get you , sorry! Did you mean to say that here is a substitution also?

hartnn · Answer

$\Large \lim \limits_{x \to a}f(x) =f(a)$

goformit100 · Answer

Refet to Tata Mc-Graw Hill Series Mathematics Book.

mathslover · Answer

Okay, I see.

So, $l_1 = 4$ and $l_2 = 0$ and $l_3 = 0/0$ 

I don't have to do direct substitution in l_3 , right?

miracrown · Answer

|dw:1402140055271:dw|