Find the limit, if it exists.

Let [tex]$f(x)=\left\{\begin{array}{l}\frac{x^2-16}{x+4} \text { if } x\ \textgreater \ 0 \\ \frac{x^2-16}{x-4} \text { if } x\ \textless \ 0\end{array}\right.$[/tex]

Find [tex]$\lim _{x \rightarrow 0^{-}} f(x)$[/tex].

A. [tex]$-4$[/tex]
B. [tex]$4$[/tex]
C. Does not exist
D. [tex]$\infty$[/tex]



Answer :

To find the limit [tex]\(\lim_{x \to 0^-} f(x)\)[/tex], we need to evaluate the function [tex]\(f(x)\)[/tex] as [tex]\(x\)[/tex] approaches 0 from the left side (i.e., from negative values of [tex]\(x\)[/tex]).

The given piecewise function is:
[tex]\[ f(x) = \begin{cases} \frac{x^2 - 16}{x + 4} & \text{if } x > 0, \\ \frac{x^2 - 16}{x - 4} & \text{if } x < 0. \end{cases} \][/tex]

Since we are interested in the limit as [tex]\(x\)[/tex] approaches 0 from the negative side, we use the part of the function that applies to [tex]\(x < 0\)[/tex]:
[tex]\[ f(x) = \frac{x^2 - 16}{x - 4}. \][/tex]

Let's first factorize the numerator [tex]\(x^2 - 16\)[/tex]:
[tex]\[ x^2 - 16 = (x - 4)(x + 4). \][/tex]

So the expression becomes:
[tex]\[ \frac{(x - 4)(x + 4)}{x - 4}. \][/tex]

Now, as long as [tex]\(x \neq 4\)[/tex], the [tex]\((x - 4)\)[/tex] terms in the numerator and the denominator can be canceled out. Thus, the simplified function is:
[tex]\[ f(x) = x + 4. \][/tex]

Now, we need to find the limit of this simplified function as [tex]\(x\)[/tex] approaches 0 from the left:
[tex]\[ \lim_{x \to 0^-} (x + 4). \][/tex]

Substitute [tex]\(x = 0\)[/tex]:
[tex]\[ \lim_{x \to 0^-} (x + 4) = 0 + 4 = 4. \][/tex]

Therefore, the limit is:
[tex]\[ \lim_{x \to 0^-} f(x) = 4. \][/tex]

Thus, the answer is [tex]\(4\)[/tex].