How could you find the limit of x+1/(x+2)(x-3) if the limit goes to -2 , I know that it's going to give us infinity and -infinity but if we take the limit from the right and left it's going to give us - over - and it's +.

from the right we take the value -1.9 which going to give us - over + * - " the values".

Question

How could you find the limit of x+1/(x+2)(x-3) if the limit goes to -2 , I know that it's going to give us infinity and -infinity but if we take the limit from the right and left it's going to give us - over - and it's +.

from the right we take the value -1.9 which going to give us - over + * - " the values".

kira_yamato · Answer

Use L'hopital's Rule.

anonymous · Answer

I need you to explain man :( please

kira_yamato · Answer

So is the expression $$x + \frac{1}{(x+2)(x+3)}?$$

kira_yamato · Answer

Hi are you there @Cutefriendzoned

anonymous · Answer

yeah man i'm here , and thank you for helping by the way , but no the expression is like this:

lim x goes to -2 (x+1)/(x+2)(x-3)

kira_yamato · Answer

That makes things easier.
Differentiate the numerator and denominator wrt x.

kira_yamato · Answer

Let L be the limit operator$$L \left( \frac{ u(x) }{ v(x) } \right) = L \left( \frac{ u'(x) }{ v'(x) } \right)$$

anonymous · Answer

Do we have to differentiate it ? He only wants us to find the limit for left and right.

I'm lost now :( what my prof said that just look at (x+2) and see if it gives you + or - if it gives you - and the numerator gives you - then it goes to +infinity.

is this correct ?

kira_yamato · Answer

You differentiate if you wanna use L'hopital's rule. Because in this case you have something indeterminate. So the only way is to apply L'hopital's rule (i.e. differentiate)

kira_yamato · Answer

$$L \left( \frac{ \color{red}{u(x)} }{ \color{green}{v(x)} } \right) = L \left( \frac{ \color{red}{u'(x)} }{ \color{green}{v'(x)} } \right)$$

kira_yamato · Answer

Trust me, it will help you quite a bit ^_^

anonymous · Answer

Oh man , sorry for being late, but it's going to be like this if we're going to do your way lim x goes to 1 which is 1 over lim x goes to -2 for 2x-7 ? 

either i'm doing something wrong or I didn't get it :(

tkhunny · Answer

A little late to this party, but do NOT use L'Hospital's Rule on this one.  It is NOT an indeterminate form a the rule does not apply.  It is the invalid and careless application of L'Hospital's rule.  Don't do it!

If you mean $ \dfrac{x+1}{(x+2)(x-3)}$, your task is only to determine the sign of anything around x = -2.  Clearly, the function increases without bound as it approaches -2.  The only question is positive or negative.

anonymous · Answer

I think that I got it , thank you.

tkhunny · Answer

* as x approaches -2.

I sort of said as the function approaches -2.  That would be at least misleading, if not simply incorrect.

kira_yamato · Answer

Thanks @tkhunny. Guess I overlooked that point on the condition of L'Hopital's

tkhunny · Answer

No worries.  L'Hospital's Rule is a pretty amazing result.  It is VERY TEMPTING to use it.

kira_yamato · Answer

I agree...

tkhunny · Answer

It is also VERY annoying when the problem statement says, "Find the limit, and DON'T use L'Hospital!"  :-)

Just make sure it's an indeterminate form before jumping in!  Sadly, it doesn't ALWAYS help.

kira_yamato · Answer

True... Those are the worst problems in limits, ever...