Question

lim
x→8
√x²+2x−3

Explanation:

Step1: Substitute \(x = 8\) into the function

We have the function \(y=\sqrt{x^{2}+2x - 3}\). When \(x = 8\), we calculate the value inside the square - root first.

$$x^{2}+2x - 3=(8)^{2}+2\times8 - 3$$

$$=64 + 16-3$$

$$=77$$

Step2: Find the limit

Since the function \(y = \sqrt{x^{2}+2x - 3}\) is continuous for \(x^{2}+2x - 3\geq0\) (and for \(x = 8\), \(x^{2}+2x - 3=77>0\)), by the property of continuous functions \(\lim_{x
ightarrow a}f(x)=f(a)\) when \(f(x)\) is continuous at \(x = a\). So \(\lim_{x
ightarrow8}\sqrt{x^{2}+2x - 3}=\sqrt{77}\)

Answer:

\(\sqrt{77}\)