Find
$$\lim _ { x \rightarrow 6 } \frac { \sqrt { x + 7 } } { x - 2 }$$
by direct substitution.
A) $\frac { \sqrt { 6 } } { 4 }$
B) $\frac { \sqrt { 13 } } { 2 }$
C) $\frac { \sqrt { 13 } } { 4 }$
D) $\frac { 1 } { 4 }$
E) $\sqrt { 13 }$

Question

Quizplus · Accepted Answer

Direct substitution means plugging in the value 6 for x:
$$\lim_{x\rightarrow6}\frac{\sqrt{x+7}}{x-2}=\frac{\sqrt{6+7}}{6-2}=\frac{\sqrt{13}}{4}$$

Find lim⁡x→6x+7x−2\lim _ { x \rightarrow 6 } \frac { \sqrt { x + 7 } } { x - 2 }x→6lim​x−2x+7​​

Find $\lim _ { x \rightarrow 6 } \frac { \sqrt { x + 7 } } { x - 2 }$