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hw5 the limit laws (target l4; §2.3)
score: 6/13 answered: 6/13
question 7
evaluate the limit: $lim_{x \to 0}\frac{sqrt{4x + 64}-8}{x}$
question help: video message instructor

Explanation:

Step1: Rationalize the numerator

Multiply the fraction by $\frac{\sqrt{4x + 64}+8}{\sqrt{4x + 64}+8}$. We get $\lim_{x
ightarrow0}\frac{(\sqrt{4x + 64}-8)(\sqrt{4x + 64}+8)}{x(\sqrt{4x + 64}+8)}$.
Using the difference - of - squares formula $(a - b)(a + b)=a^{2}-b^{2}$, the numerator becomes $(4x + 64)-64=4x$. So the limit is $\lim_{x
ightarrow0}\frac{4x}{x(\sqrt{4x + 64}+8)}$.

Step2: Simplify the fraction

Cancel out the common factor $x$ in the numerator and denominator. We have $\lim_{x
ightarrow0}\frac{4}{\sqrt{4x + 64}+8}$.

Step3: Evaluate the limit

Substitute $x = 0$ into the expression. $\frac{4}{\sqrt{4\times0+64}+8}=\frac{4}{\sqrt{64}+8}=\frac{4}{8 + 8}=\frac{4}{16}=\frac{1}{4}$.

Answer:

$\frac{1}{4}$