limit (x^2+x-6)/(x+3) when x approaches -3

first we need to find the value of the function when x=-3

then, lim (-3^2 -3 -6)/-3+3 = 0/0

0/0 means that this method has failed because both fuctions has a root of -3

Now we try to factor the function:

limit (x^2+x-6)/(x+3) = lim (x+3)(x-2)/(x+3 = lim (x-2)

then lim(x-2) as x --> -3 is (-3-2) = -5

For evaluating the limit, we'll choose the dividing out technique.

We'll apply the direct substitution, by substituting the unknown x, by the value -3 and we'll see that it fails, because both, numerator and denominator, are cancelling for x=-3. That means x=-3 is a root for both, that means that (x+3) is a common factor for both.

We'll write the numerator using the formula:

x^2+x-6=(x-x1)(x-x2), where x1, x2 are the roots and x1=-3

x^2+x-6=(x+3)(x-x2)

We also know that x1+x2 = -1, -3+x2=-1

and x1*x2=-6, (-3)*x2=-6

x2=2

Now, we'll evaluate the limit:

lim (x^2+x-6)/(x+3) = lim (x+3)(x-2)/(x+3)

Now, we can divide out like factor:

lim (x^2+x-6)/(x+3) = lim (x-2)

We can apply the replacement theorem and we'll get:

lim (x-2) = -3-2 = -5

**So, lim (x^2+x-6)/(x+3) = -5.**

To find the lt (x^2+x-6)/(x+3) as xapproaches -3.

Solution:

We see that f(x) = (x^2+x-6)/(x+3) goes for 0/0 form at x=-3.

So we try numerator. :

f(x) = (x+3)(x-2/(x+3) = x-2.

Therefore, lt x--> -3 f(x) = lt x--> 3 of (x-2 ) = (-3-2) = -5.