2
$\begingroup$

How can you prove, via first principles, that the limit $$\lim_{h\rightarrow 0}\frac{\sqrt{(1-h)^2+(2+h)^2}-\sqrt{5}}{h}$$ exists?

Somehow, I wasn't able to do it, without using specific properties of the function $\frac{\sqrt{(1-h)^2+(2+h)^2}-\sqrt{5}}{h}$ and just using basic propositions about limits.

$\endgroup$
1
$\begingroup$

Hint Multiply the numerator and denominator by $\sqrt{(1-h)^2+(2+h)^2}+\sqrt{5}$. Then you get $$\lim_{h\to 0}\frac{(1-h)^2+(2+h)^2-5}{h(\sqrt{(1-h)^2+(2+h)^2}+\sqrt{5})}=\lim_{h\to 0}\frac{1-2h+h^2+4+4h+h^2-5}{h2\sqrt 5}\\=\lim_{h\to 0}\frac{h(2+2h)}{h2\sqrt 5}=\lim_{h\to 0}\frac{2+2h}{2\sqrt 5}=\frac 1{\sqrt 5}$$

$\endgroup$
2
$\begingroup$

What happens when you multiply the numerator and denominator by the (clearly non-zero) conjugate $$\sqrt{(1-h)^2+(2+h)^2} + \sqrt{5}?$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.