Substitution Attempt: First, let's try to directly substitute the value of x into the expression to see if the limit can be evaluated this way.x→−4limx2+x−127x+28=(−4)2−4−127(−4)+28=16−4−12(−28+28)=00We get an indeterminate form 0/0, which means we need to use a different method to find the limit.
Factoring the Denominator: Since we have an indeterminate form, we should try to factor the quadratic expression in the denominator and see if there is a common factor that can be canceled with the numerator.The quadratic expression x2+x−12 can be factored into (x+4)(x−3).
Rewriting with Factored Denominator: Now let's rewrite the limit expression with the factored denominator and see if we can simplify it. limx→−4(x+4)(x−3)7x+28= limx→−4(x+4)(x−3)7(x+4)Notice that (x+4) is a common factor in both the numerator and the denominator.
Canceling Common Factor: We can now cancel the common factor (x+4) from the numerator and the denominator.x→−4lim(x+4)(x−3)7(x+4)=x→−4limx−37Now that the expression is simplified, we can directly substitute x=−4 into the limit.
Direct Substitution: Substitute x=−4 into the simplified expression to find the limit.x→−4lim(x−3)7= \frac{7}{((−4)−3)}= \frac{7}{(−7)}= −1We have found the limit, and it exists.
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