Answer to Question #191546 in Real Analysis for Amara

Question

f (x) be defined as follows

 f(x) = {x^ 2+8x+15/x+3 ifx<−3

x^2 − 7 if x ≥ −3 }

Prove from first principles (i.e.an ε − δ proof) that f is continuous at the point x = −3.

Answer 1

$f(x) = \dfrac{x^2+8x+15}{x+3}$ if, $x<-3$

$= x^2-7$ if, $x\ge -3$

$RHL = limx \rightarrow-3^{+}(x^2-7) = (-3)^2-7 = 2$

$LHL = limx \rightarrow-3^{-} \dfrac{x^2+8x+15}{x+3}$

$= limx \rightarrow -3^{-}\dfrac{(x+3)(x+5)}{(x+3)}$

$= limx \rightarrow -3^{-}(x+5)$

$= -3+5 = 2$

Also f(x) at $x = -3 = (-3)^2-7 = 9-7 = 2$

We can clearly see that,

$LHL=RHL=f(-3)$ Hence the given function is continuous at $x=-3.$