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How would you go about proving that:

$\lim_{x\to2^-}\frac{1}{x-2}= -\infty$ (i.e. x approaching 2 from left hand side)

Been struggling to make much headway on this. Not sure how I'm supposed to use Epsilon in this case. So far, I have:
Given any negative number M there exists $\delta$ s.t. $f(x) < M$ if $0<|x-2|<\delta$
-> $\frac{1}{x-2} < M$ if "..."
-> $\frac{1}{M} > x-2$ if "..."

Really not sure how to proceed from here. Can't find any online examples like it and the one in my book is for $\frac{1}{x^2}$ (I understand it, but not sure how to apply it here).

Any help/feedback/advice would be greatly appreciated.

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Let $M<0$ given.

we look for $\delta$ such that

if

$\color{red}{0<2-x<\delta}$ then

$\frac{1}{x-2}<M<0$

or

$\frac{1}{2-x}>-M>0$

which becomes after inversion

$\color{red}{0<2-x<-\frac{1}{M}}$

so we can take

$\delta=-\frac{1}{M}$ .

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Assume $2-\frac 1 n <x < 2$. Then $-\frac 1 n < x-2 < 0$ and hence $\frac 1 {x-2} < - \frac 1 {1/n} = - n$.

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  • $\begingroup$ Sorry if it seems stupid, but could you explain the thought process behind this. I think I'm just missing something $\endgroup$ – Ohmega Oct 15 '16 at 21:48
  • $\begingroup$ $n$ is large so $1/n$ is small so $x$ is between a number slightly less than $2$ and $2$. Then we subtract $2$ throughout and take reciprocals (thus reversing the order) with the result that your term is less than any negative number. $\endgroup$ – user301988 Oct 15 '16 at 23:20

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