Question

Value of lim x->2 (1/(x-2)-4/(x^2))
indeterminate inf-inf so L'Hospital rule
combine terms and I think I get
1-4/x-2-x^2-4?

Answer 1

Difficult to understand the terms.

Answer 2

\[\lim_{x \rightarrow 2} (\frac{ 1 }{ x-2 }-\frac{ 4 }{ x^2-4 })\] combine to \[\lim_{x \rightarrow 2}(\frac{ 1-4 }{ x-2-x^2-4 })\]

Answer 3

which I'm certain is wrong

Answer 4

yeah 2 nd step is wrong.

Answer 5

First write x^2-4 as (x+2)(x-2).
Then take out 1/(x-2) out of the terms.
You will be having\[1/(x-2) [ 1-4/(x+2)]\]
now you can proceed.

Answer 6

So I'm left with \[\lim_{x \rightarrow 2}(\frac{ 4 }{ x+2 })\]? That doesn't seem right or am I completely missing something?

Answer 7

no, how are you still left with the four?

Answer 8

oh so I've got -3/x+2?

Answer 9

no, 1- 4/x+2 = x+2-4/x+2
=x-2/x+2
the x-2 gets cancelled.
so all thats left is
1/x+2

Answer 10

Aaaaaa- that helps. But I'm pretty sure that the question is asking for me to apply L'Hospital's Rule which I need to show the work on

Answer 11

so the \[\lim_{x \rightarrow 2}=\frac{ 1 }{ 4 }\]

Answer 12

L'hospital method can be applied only when you have a indeterminate form.
The answer is correct.

Answer 13

Thanks

Answer 14

\[\large \lim_{x \rightarrow 2} (\frac{ 1 }{ x-2 }-\frac{ 4 }{ x^2-4 })\]evaluate the inside.\[\large \frac{1}{x-2}-\cfrac{4}{(x-2)(x+2)}\]Make a common denominator. \[\large \cfrac{(x+2)-4}{(x+2)(x-2)}\]Simplify \[\large \cfrac{x-2}{(x+2)(x-2)}\]Simplify again. \[\large \cfrac{1}{x+2}\]Now take the limit of \(\cfrac{1}{x+2}\)\[\large \lim_{x \rightarrow 2}\cfrac{1}{x+2}\]\[\large =\cfrac{1}{4} \]