a) lim ( x tends to 2 ) 1 / ( x - 2 )

b) lim (x tends to 2) 1/ (x-2)^2

I would be grateful if you can help me with this, providing clear explanations

Thanks in advance!

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- Sep 2nd 2008, 11:04 PMose90Limit of a function as x tends to real number
a) lim ( x tends to 2 ) 1 / ( x - 2 )

b) lim (x tends to 2) 1/ (x-2)^2

I would be grateful if you can help me with this, providing clear explanations

Thanks in advance! - Sep 2nd 2008, 11:11 PMProve It
- Sep 2nd 2008, 11:22 PMose90
Good explanation , thanks!

However, the answer shown in my notes for a) is no limit, is it obviously wrong? - Sep 2nd 2008, 11:32 PMMoo
Yes.

Consider x to 2, with x>2 (from the right). Then x-2>0.

So it will be**positive**infinity.

Consider x to 2, with x<2 (from the left). Then x-2<0.

So it will be**negative**infinity.

Thus $\displaystyle \lim_{\substack{x \to 2 \\ x>2}} \quad \frac{1}{x-2} \neq \lim_{\substack{x \to 2 \\ x<2}} \quad \frac{1}{x-2}$

This means the limit doesn't exist, because the limit is not the same from the right and from the left. - Sep 3rd 2008, 12:51 AMmr fantastic
It's essential that left hand and right and right hand limits are always considered.

It's not good enough to say x - 2 --> 0 etc.

You need to consider $\displaystyle x \rightarrow 2^+$ and $\displaystyle x \rightarrow 2^-$, whether the limiting value of x - 2 etc. is $\displaystyle 0^+$ or $\displaystyle 0^-$ and consequently whether the left and right handed limits are +oo or -oo .... - Sep 3rd 2008, 02:02 AMose90
Thanks everyone, I have understood.