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Which of the following is true for the validity of the limit of the function $\lim _{x \rightarrow-8} \frac{x^2+6 x-16}{x+8} ?$
Option: 1
For every point in the domain of the function and $x=-8$

Option: 2
For every point in the domain of the function except $x=-8$

Option: 3
For every point in the domain of the function only

Option: 4
None of the above

Answers (1)

Note the following points.

The domain of any function is the set of all its inputs for which the function itself remains valid.
The limit of any function $\lim _{x \rightarrow p^{+}} f(x)$ may exist even when f(x) is not defined at $x=p$

Now, the provided function $f(x)=\frac{x^2+6 x-16}{x+8}$ can be reduced as follows.

$f(x)$

$=\frac{x^2+6 x-16}{x+8}$

$=\frac{x^2+8 x-2x-16}{x+8}$

$=\frac{x(x+8)-2 (x-8)-16}{x+8}$

$=\frac{(x+8) (x-2)}{x+8}$

$=(x-2) ,$ $x \neq -8$

Evidently, the function $f(x)$ becomes invalid at $x=-8$ . So, the domain of the function $f(x)$ does not have the point $x=-8$

But the limit of the function exists as evident from the following calculation.

$\lim _{x \rightarrow-8} \frac{x^2+6 x-16}{x+8}$

$=\lim _{x \rightarrow-8} \frac{(x+8)(x-2)}{x+8}$

$=\lim _{x \rightarrow-8} (x-2)$ $[ x\neq-8]$

$=-8-2$

$=-10$

Thus, the limit of the function $\lim _{x \rightarrow-8} \frac{x^2+6 x-16}{x+8}$ exist for every point in the domain of the function and $x=-8$

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Kshitij

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