$\text { Let } f ( x ) = 3 + 2 x ^ { 2 } \text { and } g ( x ) = \sqrt { x + 3 } \text {. Find the limit. }$ $\lim _ { x \rightarrow 2 } g ( f ( x ) )$
A) $\sqrt { 6 }$
B) $\sqrt { 14 }$
C) $\sqrt { 11 }$
D) $\sqrt { 10 }$
E) $\sqrt { 2 }$

Question

Quizplus · Accepted Answer

First, substitute $f(x) = 3 + 2x^2$ into $g(x) = \sqrt{x + 3}$ to get $g(f(x)) = \sqrt{3 + 2x^2 + 3}$. Simplifying inside the square root gives $\sqrt{6 + 2x^2}$. Then, substitute $x = 2$ into this expression to get $\sqrt{6 + 2(2)^2} = \sqrt{6 + 8} = \sqrt{14}$.

Let f(x)=3+2x2 and g(x)=x+3. Find the limit. \text { Let } f ( x ) = 3 + 2 x ^ { 2 } \text { and } g ( x ) = \sqrt { x + 3 } \text {. Find the limit. } Let f(x)=3+2x2 and g(x)=x+3​. Find the limit.

$\text { Let } f ( x ) = 3 + 2 x ^ { 2 } \text { and } g ( x ) = \sqrt { x + 3 } \text {. Find the limit. }$