it: $lim_{x \to 13} \frac{x - 13}{sqrt{x + 3} - 4}$
it: $lim_{x \to 13} \frac{x - 13}{sqrt{x + 3} - 4}$
Answer
Explanation:
Step1: Rationalize the denominator
Multiply the numerator and denominator by the conjugate of the denominator (\sqrt{x + 3}+4). [ \begin{align*} \lim_{x\rightarrow13}\frac{x - 13}{\sqrt{x + 3}-4}&=\lim_{x\rightarrow13}\frac{(x - 13)(\sqrt{x + 3}+4)}{(\sqrt{x + 3}-4)(\sqrt{x + 3}+4)}\ \end{align*} ] Using the difference - of - squares formula ((a - b)(a + b)=a^{2}-b^{2}), where (a=\sqrt{x + 3}) and (b = 4), the denominator becomes ((\sqrt{x+3})^{2}-4^{2}=x + 3-16=x - 13). So the limit becomes (\lim_{x\rightarrow13}\frac{(x - 13)(\sqrt{x + 3}+4)}{x - 13}).
Step2: Simplify the expression
Cancel out the common factor ((x - 13)) (since (x\neq13) when taking the limit, we can do this cancellation). [ \begin{align*} \lim_{x\rightarrow13}\frac{(x - 13)(\sqrt{x + 3}+4)}{x - 13}&=\lim_{x\rightarrow13}(\sqrt{x + 3}+4) \end{align*} ]
Step3: Evaluate the limit
Substitute (x = 13) into (\sqrt{x + 3}+4). [ \begin{align*} \sqrt{13+3}+4&=\sqrt{16}+4\ &=4 + 4 \end{align*} ]
Answer:
(8)