Question

Evaluate the Limits?

Evaluate the limits
A) `lim_(x->2) ((3x^2-7x+2)/(x-2))`
B) `lim_(x->-3)(x+3)/(sqrt(x^2-5)-2)`
C) `lim_(x->-2)(x^3+x^2-8x-12)/(x^3+8x^2+20x+16)`

Answer 1

A) `5`

B) `-2/3`

C) `-5/2`

Explanation:

One way to do this is to graph this and look at the the function as it approaches the desired limits so in this case for the first one as `x` is nearly at `2`, `x` nearly at `-3` and `x` at nearly `-2`

If you want the more specific way then do this:

A) The first one is solved by factoring because if you directly substitute the number `2` in for the `x`'s you'll see you get `0/0` which you can't have.

Factor the numertor:

`lim_(x->2)=(((x-2)(3x-1))/(x-2))`

`lim_(x->2)=(((cancel(x-2))(3x-1))/cancel((x-2)))`

`lim_(x->2)=3x-1`

now subsitute the `2` into `x` in the equation:

`lim_(x->2) =5`

B) This one also gives `0/0` if you plug-in the `-3` straight away.

Flip the numerator and denominator

`lim_(x->-3)= (x+3)/(sqrt(x^2-5)-2)`

`lim_(x->-3)= (sqrt(x^2-5)+2)/(x-3)`

Plug-in `x=-3`

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

C) If you plug-in right away you get the same thing `0/0`

Here what you do is take the derivative of both numerator and denominator seperately because you have an `x^3` on both numerator and denominator

Derivative of numerator:

`(x^3+x^2-8x-12) d/dx`

use the power rule and you get:

`3x^2+2x-8`

Derivative of denominator:

`(x^3+8x^2+20x+16) d/dx`

using power rule again:

`3x^2+16x+20`

Now to evaluate the limit by plugging-in `x=-2`:

`lim_(x->-2)=(3x^2+2x-8)/(3x^2+16x+20)`

`lim_(x->-2)=-5/2`