Can Any one help me with this limit ?

$$\lim_{x \rightarrow 2}\frac{1}{x}\left( \frac{1}{x-2} -\frac{1}{2}\right)$$

Question

Can Any one help me with this limit ? 

$$\lim_{x \rightarrow 2}\frac{1}{x}\left( \frac{1}{x-2} -\frac{1}{2}\right)$$

unklerhaukus · Answer

$$\frac{1}{x-2}=\frac{1}{x-2}\times1$$$$=\frac{1}{x-2}\times\frac{x-2}{x-2}$$$$=\frac{x-2}{x^2-4x+4}$$
hmm that didn't help

anonymous · Answer

... I think like your way .. its not work.

unklerhaukus · Answer

$$=\frac{1}{x-2}\times\frac{x+2}{x+2}$$$$=\frac{x+2}{x^2-4}$$
damn didn't help either

anonymous · Answer

hahaha ... Its true ... I spend 30 mint by verifying these way .. you wrote it . its give true answer .

anonymous · Answer

its give true answer in solution .. but not this . way .. any way .. i don't want to lead ppl to wrong answer .. i just look4 it ^_^

unklerhaukus · Answer

$$\lim_{x\rightarrow2}\quad\frac{x-2}{x^2-4x+4}\rightarrow\frac 00$$, 
so you can use l'Hôpital's rule

anonymous · Answer

so ? does this lead for certain answer ? if i took Hopital rule does that make answer 1/4 ?

anonymous · Answer

@UnkleRhaukus  , thnx for trying :)

foolaroundmath · Answer

The limit does not exist. 1/(x-2) becomes inf when approached from right and becomes -inf when approached from left

anonymous · Answer

@FoolAroundMath , the Limit exist , and its answer 1/4 ,, i just need to verify it :)

foolaroundmath · Answer

anonymous · Answer

I want , or wish to believe it not exist ,, for that star for you , thnx

unklerhaukus · Answer

the graph is quite convincing