Find $\lim _{y \rightarrow 0} \frac{\sqrt{10+y}-\sqrt{10}}{y}$ .
A) $\frac { \sqrt { 10 } } { 20 }$
B) $\frac { \sqrt { 10 } } { 10 }$
C)0
D) $\frac { \sqrt { 10 } } { 2 }$
E)limit does not exist

Question

Quizplus · Accepted Answer

To solve this limit, we can multiply the numerator and denominator by the conjugate of the numerator, $\sqrt{10+y}+\sqrt{10}$, to eliminate the square root in the denominator. This results in:$$\lim _{y \rightarrow 0} \frac{\sqrt{10+y}-\sqrt{10}}{y} = \lim _{y \rightarrow 0} \frac{(\sqrt{10+y}-\sqrt{10})(\sqrt{10+y}+\sqrt{10})}{y(\sqrt{10+y}+\sqrt{10})}$$$$= \lim _{y \rightarrow 0} \frac{10+y-10}{y(\sqrt{10+y}+\sqrt{10})}$$$$= \lim _{y \rightarrow 0} \frac{y}{y(\sqrt{10+y}+\sqrt{10})}$$$$= \lim _{y \rightarrow 0} \frac{1}{\sqrt{10+y}+\sqrt{10}}$$$$= \frac{1}{2\sqrt{10}}$$$$= \frac{\sqrt{10}}{2\sqrt{10}\sqrt{10}}$$$$= \frac{\sqrt{10}}{20}$$Therefore, the correct answer is $\frac{\sqrt{10}}{20}$, which is choice A.

Find lim⁡y→010+y−10y\lim _{y \rightarrow 0} \frac{\sqrt{10+y}-\sqrt{10}}{y}limy→0​y10+y​−10​​ A) 1020\frac { \sqrt { 10 } } { 20 }2010​​

Find $\lim _{y \rightarrow 0} \frac{\sqrt{10+y}-\sqrt{10}}{y}$
A) $\frac { \sqrt { 10 } } { 20 }$