Write down an expression in the form $ax^n$ for: $\lim_{h\to 0} \frac{\sqrt{x+h}-\sqrt{x}}{h}$

Question

Write down an expression in the form $ax^n$ for

$$\lim_{h\to 0} \frac{\sqrt{x+h}-\sqrt{x}}{h}$$

What I have tried so far:

multiplying by the conjugate to give:

$$\lim_{h\to 0} \frac{\sqrt{x+h}-\sqrt{x}}{h} \cdot \frac{h}{\sqrt{x+h}+\sqrt{x}}$$

so we cancel out $h$:

$$\lim_{h\to 0} \frac{\sqrt{x+h}-\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}$$

as $\lim_{h\to 0}$:

$$\lim_{h\to 0} \frac{\sqrt{x+0}-\sqrt{x}}{\sqrt{x+0}+\sqrt{x}}$$

giving:

$$\lim_{h\to 0} \frac{\sqrt{x}-\sqrt{x}}{\sqrt{x}+\sqrt{x}}$$

So I'm not sure how you get the numerator to equal $1$ which would give the correct given answer: $0.5x^{-0.5}$

I think I'm missing a simpler method to obtain this answer - any hints would be much appreciated!

Answer 1

Using the concept of derivatives: We know that: $$f'(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$$ Comparing what we have in the question to this formula, it will be clear to us that what is being asked is essentially: $$\frac{\mathrm{d}}{\mathrm{d}x}\sqrt{x}$$ Hence we will finally end up with: $$\boxed{0.5x^{-0.5}}$$ And we're done!

Answer 2

$\frac {\sqrt {x+h} -\sqrt x }h =\frac {(x+h)-x} {h(\sqrt {x+h} +\sqrt x)}=\frac 1 {\sqrt {x+h} +\sqrt x}\to \frac 1 {2\sqrt x}=\frac1 2 x^{-1/2}$.

Answer 3

$$\lim_{h\to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}$$ $$=\lim_{h\to 0}\frac{(\sqrt{x+h}-\sqrt{x})(\sqrt{x+h}+\sqrt{x})}{h(\sqrt{x+h}+\sqrt{x})}$$ Use identity:$(a-b)(a+b)=a^2-b^2$ $$=\lim_{h\to 0}\frac{(\sqrt{x+h})^2-(\sqrt{x})^2}{h(\sqrt{x+h}+\sqrt{x})}$$ $$=\lim_{h\to 0}\frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})}$$ $$=\lim_{h\to 0}\frac{1}{\sqrt{x+h}+\sqrt{x}}=\frac{1}{2\sqrt{x}}=0.5x^{-0.5}$$