Question

find the limit$\underset{x\to 3}{\lim }\sqrt{x^2+7}$

Short answer

$\underset{x\to 3}{\lim }\sqrt{x^2+7}=4$

Step-by-step solution

Step 1. Given information

$\underset{x\to 3}{\lim }\sqrt{x^2+7}$

Step 2.  Limit of the function  

We can find limit of root function by $\underset{x\to c}{\lim }f\left(x\right)=f\left(c\right)$if c is in the domain of f(x) and the domain of f(x)=$\sqrt{x^2+7}$is all real numbers.

So we can use the above formula .

$$\begin{gathered} \underset{x\to 3}{\lim }f\left(x\right)=f\left(3\right) \\ =\sqrt{(3{)}^2+7}=4 \end{gathered}$$