details\nno additional details were added for this assignment.\nhw5 the limit laws (target l4; §2.3)\nscore…

details\nno additional details were added for this assignment.\nhw5 the limit laws (target l4; §2.3)\nscore: 6/13 answered: 6/13\nquestion 7\nevaluate the limit: $lim_{x \to 0}\frac{sqrt{4x + 64}-8}{x}$\nquestion help: video message instructor
Answer
Explanation:
Step1: Rationalize the numerator
Multiply the fraction by $\frac{\sqrt{4x + 64}+8}{\sqrt{4x + 64}+8}$. We get $\lim_{x\rightarrow0}\frac{(\sqrt{4x + 64}-8)(\sqrt{4x + 64}+8)}{x(\sqrt{4x + 64}+8)}$. 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 $\lim_{x\rightarrow0}\frac{4x}{x(\sqrt{4x + 64}+8)}$.
Step2: Simplify the fraction
Cancel out the common factor $x$ in the numerator and denominator. We have $\lim_{x\rightarrow0}\frac{4}{\sqrt{4x + 64}+8}$.
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:
$\frac{1}{4}$