evaluate the limit\n\n$$\\lim_{x \\to \\infty} \\sqrt{x^{2}+4x + 20}-x$$
evaluate the limit\n\n$$\\lim_{x \\to \\infty} \\sqrt{x^{2}+4x + 20}-x$$
Answer
Explanation:
Step1: Rationalize the expression
Multiply the numerator and denominator by the conjugate of (\sqrt{x^{2}+4x + 20}-x), which is (\sqrt{x^{2}+4x + 20}+x). [ \begin{align*} \lim_{x\rightarrow\infty}(\sqrt{x^{2}+4x + 20}-x)&=\lim_{x\rightarrow\infty}\frac{(\sqrt{x^{2}+4x + 20}-x)(\sqrt{x^{2}+4x + 20}+x)}{\sqrt{x^{2}+4x + 20}+x}\ &=\lim_{x\rightarrow\infty}\frac{(x^{2}+4x + 20)-x^{2}}{\sqrt{x^{2}+4x + 20}+x}\ &=\lim_{x\rightarrow\infty}\frac{4x+20}{\sqrt{x^{2}+4x + 20}+x} \end{align*} ]
Step2: Divide numerator and denominator by (x)
Since (x\rightarrow\infty), (x>0). Divide each term in the numerator and denominator by (x). [ \begin{align*} \lim_{x\rightarrow\infty}\frac{4x + 20}{\sqrt{x^{2}+4x + 20}+x}&=\lim_{x\rightarrow\infty}\frac{4+\frac{20}{x}}{\sqrt{1+\frac{4}{x}+\frac{20}{x^{2}}}+1} \end{align*} ]
Step3: Apply the limit
Use the limit rule (\lim_{x\rightarrow\infty}\frac{a}{x^{n}} = 0) for (n>0) and (a) is a constant. [ \begin{align*} \lim_{x\rightarrow\infty}\frac{4+\frac{20}{x}}{\sqrt{1+\frac{4}{x}+\frac{20}{x^{2}}}+1}&=\frac{4 + 0}{\sqrt{1+0+0}+1}\ &=\frac{4}{2} \end{align*} ]
Answer:
(2)