Question

Lim x−→2 (x^2 − 4) tan(2x − 4) / (x − 2)^2 Find the following limits. Show all steps. If the limit does not exist, explain why not? Not allow to use L’Hospitals rule.

Answer 1

`lim_(x->2) ((x^2-4)tan(2x-4))/(x-2)^2 = 8`

Explanation:

Evaluate the limit:

`lim_(x->2) ((x^2-4)tan(2x-4))/(x-2)^2`

Note that:

`(x^2-4) = (x+2)(x-2)`

so we can simplify:

`lim_(x->2) ((x^2-4)tan(2x-4))/(x-2)^2 = lim_(x->2) ((x+2)(x-2)tan(2x-4))/(x-2)^2 = lim_(x->2) ((x+2)tan(2x-4))/(x-2)`

Multiply numerator and denominator by `2`:

`lim_(x->2) ((x^2-4)tan(2x-4))/(x-2)^2 = lim_(x->2) (2(x+2)tan(2x-4))/(2x-4)`

Now consider the limit:

`lim_(x->2) tan(2x-4)/(2x-4)`

Substituting `y=2x-4`, as `lim_(x->2) 2x-4 = 0` we have:

`lim_(x->2) tan(2x-4)/(2x-4) = lim_(y->0) tany/y`

that is a well known trigonometric limit:

`lim_(y->0) tany/y = lim_(y->0) 1/cosy siny/y = 1`

So:

`lim_(x->2) ((x^2-4)tan(2x-4))/(x-2)^2 = lim_(x->2) 2(x+2) * lim_(x->2) tan(2x-4)/(2x-4) = 8*1 = 8`

graph{((x^2-4)tan(2x-4))/(x-2)^2 [-4.873, 5.127, 6.26, 11.26]}