Find $\lim _ { x \rightarrow 6 } \frac { \sqrt { x + 5 } } { x - 2 }$ by direct substitution.
A) $\frac { \sqrt { 11 } } { 4 }$
B) $\frac { \sqrt { 11 } } { 2 }$
C) $\frac { \sqrt { 6 } } { 4 }$
D) $\sqrt { 11 }$
E) $\frac { 1 } { 4 }$

Question

Quizplus · Accepted Answer

By directly substituting $x = 6$ into the given limit, we get $\frac{\sqrt{6 + 5}}{6 - 2} = \frac{\sqrt{11}}{4}$.

Find lim⁡x→6x+5x−2\lim _ { x \rightarrow 6 } \frac { \sqrt { x + 5 } } { x - 2 }limx→6​x−2x+5​​

Find $\lim _ { x \rightarrow 6 } \frac { \sqrt { x + 5 } } { x - 2 }$