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!
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.
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 ....