`lim_(x->-1) (x^2 + 2x + 1)/(x^4 - 1)`

sol:

`lim_(x->-1) (x^2 + 2x + 1)/(x^4 - 1)`

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

=> `lim_(x->-1) (x+ 1)^2/((x^2 - 1^2)(x^2+1^2))`

=> `lim_(x->-1) (x+ 1)^2/((x - 1)(x+1)(x^2+1^2))`

=> `lim_(x->-1) (x+ 1)/((x - 1)(x^2+1^2))`

Now as `x-> -1` we get

`lim_(x->-1) (x+ 1)/((x - 1)(x^2+1^2))`

= > `lim_(x->-1) (-1+ 1)/((-1 - 1)((-1)^2+1^2)) = 0`

is the answer

Plugging in x = -1 into the function results in the indeterminate form 0/0. We could factor the numerator and expand the denominator to cancel terms and find the limit. However, since the numerator and denominator are both differentiable functions, we can apply L'Hopital's rule, which will be easier. Take the derivative of the numerator and denominator.

`lim_(x->-1) (2x + 2)/(4x^3)` Plugging in x = -1 shows that the limit is 0.