QuestionFind the limit: $\lim _{x \rightarrow 0} \frac{-3 x}{\sqrt{3 x+64}-8}$ .

Studdy Solution

STEP 1

Assumptions1. We are given the limit problem $\lim{x \rightarrow0} \frac{-3 x}{\sqrt{3 x+64}-8}$ . . We need to find the value of the limit as $x$ approaches $0$ .
3. The limit exists.

STEP 2

First, we try to substitute $x =0$ into the expression. $\lim{x \rightarrow0} \frac{- x}{\sqrt{ x+64}-8} = \frac{-(0)}{\sqrt{(0)+64}-8}$

STEP 3

implify the expression.
$\frac{-3(0)}{\sqrt{3(0)+64}-8} = \frac{0}{\sqrt{64}-8} = \frac{0}{8-8}$

STEP 4

We see that this results in a $0/0$ form, which is an indeterminate form. This means we can't determine the limit by direct substitution. We need to use another method.

STEP 5

We can use the conjugate method to simplify the expression. The conjugate of the denominator $\sqrt{3x+64}-8$ is $\sqrt{3x+64}+8$ . We multiply the numerator and denominator by this conjugate.
$\lim{x \rightarrow0} \frac{-3 x(\sqrt{3 x+64}+8)}{(\sqrt{3 x+64}-8)(\sqrt{3 x+64}+8)}$

STEP 6

implify the denominator using the difference of squares formula $a^2 - b^2 = (a-b)(a+b)$ .
$\lim{x \rightarrow0} \frac{-3 x(\sqrt{3 x+64}+8)}{(3x+64)-8^2}$

STEP 7

implify the denominator further.
$\lim{x \rightarrow0} \frac{-3 x(\sqrt{3 x+64}+)}{3x+64-64}$

STEP 8

implify the denominator to get.
$\lim{x \rightarrow0} \frac{-3 x(\sqrt{3 x+64}+8)}{3x}$

STEP 9

Cancel the common factor $3x$ from the numerator and denominator.
$\lim{x \rightarrow} -(\sqrt{3 x+64}+8)$

STEP 10

Now, substitute $x =0$ into the simplified expression.
$-\sqrt{3(0)+64}+8 = -\sqrt{64}+8 = -8+8$

STEP 11

implify the expression to get the limit.
$-8+8 =0$ So, $\lim{x \rightarrow0} \frac{-3 x}{\sqrt{3 x+64}-8} =0$ .

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