f(x) = {[(x^2-25)/(x-5), x ≠ 5][k,x=5] at x = 5

Question

Find the value of the constant k so that the given function is continuous at the indicated point : 

 \(f(x) = \begin{cases}\frac{x^2-25}{x-5}&, \quad x≠5\\k &, \quad x=5\\\end{cases} \) at x = 5

Answer 1

Given :

f(x) is continuous at x = 5 & f(5) = k

If f(x) to be continuous at x = 5,then 

f(5)^–= f(5)⁺ = f(5)

Since,

f(x) is continuous at x = 5 & f(5) = k

k = 10