Question

How do you prove that `lim_(x->1)1/(x-1)` doesnot exist using limit definition?

How do you prove that `lim_(x->1)1/(x-1)` doesnot exist using limit definition?

Answer 1

You approach the limit from both sides.

Explanation:

(1)
`x->1` from above: we take `x=1+epsilon`, where `epsilon` is getting smaller and smaller. The function will be:
`1/(cancel1+epsilon-cancel1)=1/epsilon`
As `epsilon` gets smaller the function gets larger, or:
`lim_(epsilon->0) 1/epsilon=+oo`

(2)
`x->1` from below: we take `x=1-epsilon`, where `epsilon` is getting smaller and smaller. The function will be:
`1/(cancel1-epsilon-cancel1)=-1/epsilon`
As `epsilon` gets smaller the function gets negatively larger, or:
`lim_(epsilon->0) -1/epsilon=-oo`

This means there is not one limit.
graph{1/(x-1) [-12.66, 12.65, -6.33, 6.33]}