Question

If f (1) = 1, f' (1) = 2, then write the value of $\underset{x\to 1}{\lim }\frac{\sqrt{f\left(x\right)}-1}{\sqrt{x}-1}$.

Answer 1

$$\begin{gathered} \underset{x\to 1}{\lim }\frac{\sqrt{f\left(x\right)}-1}{\sqrt{x}-1} \\ =\underset{x\to 1}{\lim }\frac{\sqrt{f\left(x\right)}-1}{\sqrt{x}-1}\times \frac{\sqrt{f\left(x\right)}+1}{\sqrt{f\left(x\right)}+1}\times \frac{\sqrt{x}+1}{\sqrt{x}+1} \\ =\underset{x\to 1}{\lim }\frac{\left(f\left(x\right)-1\right)\left(\sqrt{x}+1\right)}{\left(x-1\right)\left(\sqrt{f\left(x\right)}+1\right)} \\ =\underset{x\to 1}{\lim }\frac{f\left(x\right)-1}{x-1}\times \underset{x\to 1}{\lim }\frac{\left(\sqrt{x}+1\right)}{\left(\sqrt{f\left(x\right)}+1\right)} \\ =f\text{'}\left(1\right)\times \frac{1+1}{\sqrt{f\left(1\right)}+1} \\ =2\times \frac{2}{1+1} \\ =2 \end{gathered}$$