# Use continuity to evaluate the limit \lim_{x\to4}\frac{12+\sqrt{x}}{\sqrt

Use continuity to evaluate the limit
$\underset{x\to 4}{lim}\frac{12+\sqrt{x}}{\sqrt{12+x}}$
To find:
a) Domain of Numerator of function
b) Domain of denominator of the function
c) limit
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a) Consider a given limit:
$L=\underset{x\to 4}{lim}\frac{12+\sqrt{x}}{\sqrt{12+x}}$
Here
Numerator is $12+\sqrt{x}$ and denominator $=\sqrt{12+x}$
Now, find the domain of the numeratot.
Numerator $12+\sqrt{x}$ is defined for all $x\in \left[0,\mathrm{\infty }\right)$.
Hence domain of the numerator is $\left[0,\mathrm{\infty }\right]$
b) Now,
Denominator $\sqrt{12+x}$ is not defined for all x=-12. Hence, the domain of the denominator $\sqrt{12+x}$ is $\left(-12,\mathrm{\infty }\right)$.
Hence the domain of the function $\frac{12+\sqrt{x}}{\sqrt{12+x}}$ is $\left[0,\mathrm{\infty }\right)$
c) Further, find the limit $\underset{x\to 4}{lim}\frac{12+\sqrt{x}}{\sqrt{12+x}}$
Now,
$L=\underset{x\to 4}{lim}\frac{12+\sqrt{x}}{\sqrt{12+x}}$
$=\frac{12+\sqrt{4}}{\sqrt{12+4}}$
$=\frac{12+2}{\sqrt{16}}$
$=\frac{14}{4}$
$=\frac{7}{2}$
$⇒\underset{x\to 4}{lim}\frac{12+\sqrt{x}}{\sqrt{12+x}}=\frac{7}{2}$