Answer
L=-4, 𝛿 = 0.05
Work Step by Step
L=$\lim\limits_{x \to -5} \frac{x^{2} +6x+5}{x+5}$ = $\lim\limits_{x \to -5}\frac{(x+5)(x+1)}{x+5}$
=$\lim\limits_{x \to -5} x+1$ =-5+1 = -4
Solve the inequality |f(x) -L| < e
In the above calculation we concluded = $ \frac{x^{2} +6x+5}{x+5}$ = x+1
|x+1 +4 | <0.05
|x+5| <0.05
-0.050;
Now we need to find the value of 𝛿>0 that places the open interval
(-5-𝛿, -5+𝛿) centered at -5 inside the interval(-5.05, -4.95). distance between -5 and -5.05 is 0.05 and distance between -5 and -4.95 is 0.05. so we will take the smaller value for 𝛿, so 𝛿 = 0.05
