Use continuity to evaluate the limit limx→412+x12+x To find: a) Domain of Numerator of function b) Domain of denominator of the function c) limit

a) Consider a given limit: L=limx→412+x12+x Here Numerator is 12+x and denominator =12+x Now, find the domain of the numeratot. Numerator 12+x is defined for all x∈[0,∞). Hence domain of the numerator is [0,∞] b) Now, Denominator 12+x is not defined for all x=-12. Hence, the domain of the denominator 12+x is (−12,∞). Hence the domain of the function 12+x12+x is [0,∞) c) Further, find the limit limx→412+x12+x Now, L=limx→412+x12+x =12+412+4 =12+216 =144 =72 ⇒limx→412+x12+x=72

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Answered question

2021-10-21

Use continuity to evaluate the limit

lim_{x \to 4} \frac{12 + \sqrt{x}}{\sqrt{12 + x}}

To find:
a) Domain of Numerator of function
b) Domain of denominator of the function
c) limit

Answer & Explanation

liingliing8

Skilled2021-10-22Added 95 answers

a) Consider a given limit:

L = lim_{x \to 4} \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 [0, \infty)

.
Hence domain of the numerator is

[0, \infty]

b) Now,
Denominator

\sqrt{12 + x}

is not defined for all x=-12. Hence, the domain of the denominator

\sqrt{12 + x}

(- 12, \infty)

.
Hence the domain of the function

\frac{12 + \sqrt{x}}{\sqrt{12 + x}}

[0, \infty)

c) Further, find the limit

lim_{x \to 4} \frac{12 + \sqrt{x}}{\sqrt{12 + x}}

Now,

L = lim_{x \to 4} \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}

\Rightarrow lim_{x \to 4} \frac{12 + \sqrt{x}}{\sqrt{12 + x}} = \frac{7}{2}

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