Question

Why is the limit as `x->2` for `3/(2-x)` undefined?

Answer 1

You can use the properties of logarithms to check. Starting with `lim_(x->2) e^(3/(2-x))`:

`ln(lim_(x->2) e^(3/(2-x))) = lim_(x->2) ln(e^(3/(2-x)))`

`= lim_(x->2) 3/(2-x)`

But this function is of the form `1/(x)`, so let us examine the limit from both sides...

`=> lim_(x->2^+) 3/(2-x) = -oo`
`=> lim_(x->2^-) 3/(2-x) = oo`

You can see that here:

graph{3/(2-x) [-5.46, 14.54, -5.12, 4.88]}

Thus, undoing the `"ln"`:

`e^(lim_(x->2^+) 3/(2-x)) = e^(-oo) = 0`
`e^(lim_(x->2^-) 3/(2-x)) = e^(oo) = oo`

i.e. the limit from both sides at once is undefined.