1.

Prove that   \(f(x) = \begin{cases} \frac {x^2-x-6}{x-3}, & \quad \text{when x≠3;} \text{}\\ 5, & \quad \text{when x=3} \end{cases}\)  is continuous at x = 3 5, when x = 3

Answer»

LHL: \(\lim\limits_{x \to3^-} \)f(x) =  \(\lim\limits_{x \to3^-} \)  \(\frac{​​​​x^2-x-6}{x-3}\) 

 = \(\lim\limits_{x \to3^-} \)   \(\frac{​​​(​x+2)(x-3)}{x-3}\)   [By middle term splitting]

  = \(\lim\limits_{x \to3^-} \) x + 2

= 5 

RHL: \(\lim\limits_{x \to3^-} \)f(x) =  \(\lim\limits_{x \to3^-} \)  \(\frac{​​​​x^2-x-6}{x-3}\) 

 = \(\lim\limits_{x \to3^-} \)   \(\frac{​​​(​x+2)(x-3)}{x-3}\)   [By middle term splitting]

 = \(\lim\limits_{x \to3^-} \) x + 2

 = 5

f(3) = 5 

Since, \(\lim\limits_{x \to3} \) f(x) = f(3)

f is continuous at x=3.



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