Prove that ƒ(x) = {x^2 - 25/x-5 , when x ≠ 5 ,10, when x = 5 is continuous at x = 5 10, when x = 5

Question

Prove that   \(f(x) = \begin{cases} \frac {x^2-25}{x}, & \quad \text{when x≠5;} \text{}\\ 10, & \quad \text{when x=5} \end{cases}\)  is continuous at x = 5

Answer 1

LHL: = \(\lim\limits_{x \to5^-} \)f(x) =  \(\lim\limits_{x \to5^-} \) \(\frac{x^2 - 25}{x-5} \)

 = \(\lim\limits_{x \to5^-} \)  \(\frac{(x+5)(x-5)}{x-5} \)[By middle term splitting]

 = \(\lim\limits_{x \to5^-} \)  x + 5

= 10

RHL: = \(\lim\limits_{x \to5^+} \)f(x) =  \(\lim\limits_{x \to5^+} \) \(\frac{x^2 - 25}{x-5} \)

 = \(\lim\limits_{x \to5^+} \)  \(\frac{(x+5)(x-5)}{x-5} \)[By middle term splitting]

 = \(\lim\limits_{x \to5^+} \)  x + 5

= 10 

f(5)= 10 

Since, = \(\lim\limits_{x \to5} \) f(x) = f(5)

f is continuous at x=5.