Find the limit as x -> -2 of f(x) = (x^2 - 3x + a) / (x^2 + x - 2)

Question

Find the limit as x --> -2 of f(x) = (x² - 3x + a)/(x² + x - 2)

 

I've never see a limit problem before where there are two variables: in this example we have 'x' AND 'a'. This question was on a standardized test, and I think I bombed it. My solution was to find the value of 'a' so that I could compute the limit. I concluded that a = -10, which results in a limit as x --> -2 that is 7/3. However, my score was provided as earning 1 out of 4 possible points for my solution. What did I do wrong, and what is the correct way to do this problem?

Answer 1

It looks like you assumed a value for "a" that allows you to factor the numerator and to cancel a factor of x + 2 in the denominator, to keep the denominator from becoming zero.

 

Using a = -10,

 

(x² - 3x - 10)/(x² + x - 2) = [(x - 5)(x + 2)]/[(x - 1)(x + 2)] = (x - 5)/(x - 1), for x ≠ -2

 

If you let x = -2, then

 

(x - 5)/(x - 1) = (-2 - 5)(-2 - 1) = -7/-3 = 7/3

 

Your answer is correct if a = -10.  For another way to find the limit when a = -10 and x→-2, note that both the numerator and the denominator of the original expression are zero at x = -2.  That gives you an indeterminate form at x = -2, so we can use L'Hopital's rule to find the limit.  Are you familiar with this rule?

 

As x → -2,

 

(x² - 3x - 10) / (x² + x - 2) → (2x - 3)/(2x + 1) = -7/-3 = 7/3 at x = -2

 

Is there more information, like were you told the expression was indeterminate at x = -2? If the expression is given as indeterminate at x = -2, you do not need to know "a", because it drops out of the problem when you apply L'Hopital's rule.