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Find the Limit by Direct Substitution limx15x+1x10\lim _ { x \rightarrow 15 } \frac { \sqrt { x + 1 } } { x - 10 }

Question 61

Multiple Choice

Find the limit by direct substitution. limx15x+1x10\lim _ { x \rightarrow 15 } \frac { \sqrt { x + 1 } } { x - 10 }


A) limx15x+1x10=54\lim _ { x \rightarrow 15 } \frac { \sqrt { x + 1 } } { x - 10 } = - \frac { 5 } { 4 }
B) limx15x+1x10=54\lim _ { x \rightarrow 15 } \frac { \sqrt { x + 1 } } { x - 10 } = \frac { 5 } { 4 }
C) limx15x+1x10\lim _ { x \rightarrow 15 } \frac { \sqrt { x + 1 } } { x - 10 } = \infty
D) limx15x+1x10=45\lim _ { x \rightarrow 15 } \frac { \sqrt { x + 1 } } { x - 10 } = \frac { 4 } { 5 }
E) limx15x+1x10=45\lim _ { x \rightarrow 15 } \frac { \sqrt { x + 1 } } { x - 10 } = - \frac { 4 } { 5 }

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