Find $\lim _ { y \rightarrow 0 } \frac { \sqrt { 6 + y } - \sqrt { 6 } } { y }$ .
A) $\frac { \sqrt { 6 } } { 6 }$
B)0
C) $\frac { \sqrt { 6 } } { 2 }$
D)limit does not exist
E) $\frac { \sqrt { 6 } } { 12 }$

Question

Quizplus · Accepted Answer

To solve this limit, we can multiply the numerator and denominator by the conjugate of the numerator, $\sqrt{6+y} + \sqrt{6}$, to eliminate the square root in the numerator. This simplifies to $\frac{y}{y(\sqrt{6+y} + \sqrt{6})}$, which further simplifies to $\frac{1}{\sqrt{6+y} + \sqrt{6}}$. Taking the limit as $y$ approaches 0 gives $\frac{1}{2\sqrt{6}} = \frac{\sqrt{6}}{12}$.

Find lim⁡y→06+y−6y\lim _ { y \rightarrow 0 } \frac { \sqrt { 6 + y } - \sqrt { 6 } } { y }limy→0​y6+y​−6​​

Find $\lim _ { y \rightarrow 0 } \frac { \sqrt { 6 + y } - \sqrt { 6 } } { y }$