evaluate the following limit. use lhopitals rule when it is convenient and applicable. lim (√(x + 6) - √(x +…

evaluate the following limit. use lhopitals rule when it is convenient and applicable. lim (√(x + 6) - √(x + 8)) as x→∞ what is the most efficient way for this limit to be evaluated? select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. the limit can be evaluated by direct substitution. b. use the substitution t = 1/x and then lhopitals rule to rewrite the limit as lim as t→0 (). c. use lhopitals rule directly to rewrite the limit as lim as x→∞ (). d. manipulate the given expression algebraically to rewrite the limit as lim as x→∞ (). (simplify your answer.) evaluate the limit. lim (√(x + 6) - √(x + 8)) = (type an exact answer.) as x→∞
Answer
Explanation:
Step1: Rationalize the expression
Multiply and divide by $\sqrt{x + 6}+\sqrt{x + 8}$: [ \begin{align*} \lim_{x\rightarrow\infty}(\sqrt{x + 6}-\sqrt{x + 8})&=\lim_{x\rightarrow\infty}\frac{(\sqrt{x + 6}-\sqrt{x + 8})(\sqrt{x + 6}+\sqrt{x + 8})}{\sqrt{x + 6}+\sqrt{x + 8}}\ &=\lim_{x\rightarrow\infty}\frac{(x + 6)-(x + 8)}{\sqrt{x + 6}+\sqrt{x + 8}}\ &=\lim_{x\rightarrow\infty}\frac{x + 6 - x-8}{\sqrt{x + 6}+\sqrt{x + 8}}\ &=\lim_{x\rightarrow\infty}\frac{- 2}{\sqrt{x + 6}+\sqrt{x + 8}} \end{align*} ]
Step2: Evaluate the limit
As $x\rightarrow\infty$, we have: [ \begin{align*} \lim_{x\rightarrow\infty}\frac{-2}{\sqrt{x + 6}+\sqrt{x + 8}}&=\frac{-2}{\infty+\infty}\ &=0 \end{align*} ] The most - efficient way is to manipulate the given expression algebraically. So for the first part, the answer is D. $\lim_{x\rightarrow\infty}\frac{-2}{\sqrt{x + 6}+\sqrt{x + 8}}$
Answer:
D. $\lim_{x\rightarrow\infty}\frac{-2}{\sqrt{x + 6}+\sqrt{x + 8}}$ 0