Question

If the function f is continuous at x=1, find the value of k. (function in reply)

Answer 1

I'm tempted to "rationalize" the numerator. Like to give it a try?

Answer 2

How 'bout aplying l'Hôpital for the lulz?

Answer 3

Take the limit of the function definition for all \(x\ne1\). Note we have to use algebraic manipulation to turn our expression into a form defined at \(x=1\).$$\lim_{x\to1}\frac{\sqrt{x+3}-\sqrt{3x+1}}{x-1}=\lim_{x\to1}\frac{x+3-3x-1}{(x-1)(\sqrt{x+3}+\sqrt{3x+1})}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\lim_{x\to1}\frac{-2x+2}{(x-1)(\sqrt{x+3}+\sqrt{3x+1}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\lim_{x\to1}\frac{-2(x-1)}{(x-1)(\sqrt{x+3}+\sqrt{3x+1})}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\lim_{x\to1}-\frac{2}{\sqrt{x+3}+\sqrt{3x+1}}=\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\frac2{\sqrt4+\sqrt4}=-\frac24=-\frac12$$