derivative using the definition

Question

find $$f'(0) $$if$$ f(x)=\sqrt{x+\sqrt{1+x}}$$ so you set up the limit to find the derivative at a given point a $$\lim\limits_{x \to a } \frac{f(x)-f(a)}{x-a}$$ $$\lim\limits_{x \to 0} \frac{\sqrt{x+\sqrt{1+x}}-1}{x}$$ I thought of multiplying by the conjugate, but I can't get rid of the x on the denominator without causing another x to be on a denominator

Answer 1

$$\lim_{x\rightarrow0}\frac{\sqrt{x+\sqrt{x+1}}-1}{x}=\lim_{x\rightarrow0}\frac{x+\sqrt{x+1}-1}{x(\sqrt{x+\sqrt{x+1}}+1)}=\lim_{x\rightarrow0}\frac{x+1-(x-1)^{2}}{x(\sqrt{x+\sqrt{x+1}}+1)(\sqrt{x+1}-x+1)}$$ $$=\lim_{x\rightarrow0}\frac{-x^{2}+3x}{x(\sqrt{x+\sqrt{x+1}}+1)(\sqrt{x+1}-x+1)}=\lim_{x\rightarrow0}\frac{-x+3}{(\sqrt{x+\sqrt{x+1}}+1)(\sqrt{x+1}-x+1)}=\frac{3}{4}.$$

Answer 2

Let $\sqrt{1+x}=y,x=?$

Now rationalize the numerator of

$\lim_{y\to1}\dfrac{\sqrt{y^2-1+y}-1}{y^2-1}$ and cancel out $y-1$ as $y-1\ne0$ as $y\to1$