Question
If $$f\left( x \right) = \frac{x}{{\sqrt {x + 1} - \sqrt x }}$$ be a real-valued function then :
A.
$$f\left( x \right)$$ is continuous, but $$f'\left( 0 \right)$$ does not exist
B.
$$f\left( x \right)$$ is differentiable at $$x=0$$
C.
$$f\left( x \right)$$ is not continuous at $$x=0$$
D.
$$f\left( x \right)$$ is not differentiable at $$x=0$$
Answer :
$$f\left( x \right)$$ is differentiable at $$x=0$$
Solution :
$$f\left( x \right) = \frac{{x\left( {\sqrt {{x^2} + 1} + \sqrt x } \right)}}{{x + 1 - x}} = x\left( {\sqrt {{x^2} + 1} + \sqrt x } \right).$$ As $$f\left( x \right)$$ is real valued, $$x \geqslant 0.$$
$$\therefore \,f\left( x \right)$$ is differentiable at $$x=0$$ if $$\mathop {\lim }\limits_{h \to 0} \frac{{f\left( {0 + h} \right) - f\left( 0 \right)}}{h}$$ is finite and definite.
$$\mathop {\lim }\limits_{h \to 0} \frac{{f\left( {0 + h} \right) - f\left( 0 \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{h\left( {\sqrt {{h^2} + 1} + \sqrt h } \right)}}{h} = 1.$$ So, (B) is correct.
A function which is finitely differentiable is also continuous.