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Evaluate the limit assuming that $$\lim _{x \rightarrow-4} f(x)=3$$ and $$\lim _{x \rightarrow-4} g(x)=1$$

$$\lim _{x \rightarrow-4} \frac{f(x)+1}{3 g(x)-9}$$

$-\frac{2}{3}$

Calculus 1 / AB

Chapter 2

LIMITS

Section 3

Basic Limit Laws

Limits

Derivatives

Continuous Functions

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So we're given that the limit of FX as X approaches negative four equals three and the limit of G vex as X approaches negative four equals one were asked to find the limit of F of X plus one over three G of X minus nine as its approach is negative. For To do this, we can just apply the basic limit laws. First, we can apply the quotient law so we actually get the limit off after the ax plus one as ex approach is negative for and then divided by the denominator, which is the limit of three G vex minus nine as X approaches for then we can actually separate this further, using the some water and the constable to lock first. Listening to some law, you can separate this into the limit of Catholics as X approach is negative for plus the limit of one as X approaches negative for and then in the denominator, we can change this too. The limit of three g vex as ex approach is negative, for which, according to the constant multiple law, is the same as three times the limit of G of X as X approaches negative for minus the limit of nine as X approaches negative for so now all we have to do is plug in values. We know that the limit of FX as mass approaches negative four is three. So in the numerator we have three plus and we also know that the limit of one as X approaches any value will be one because ex is not in the function. So we have three plus one divided by we know that the limit of G vex as X approaches negative for is one because they were given that value. So we have three times one minus and the limit of constant is that constant. So we get nine that is equal to four over three minus nine, which is negative six. So we get negative 2/3 as our answer.

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