Find the limit $\lim_{h\to 0}\frac{ \sqrt{7(a+h)}-\sqrt{7a} }{h} $ in terms of $a$ .

Question

Find the limit in terms of the constant $a$ : $$ \lim_{h \to 0}\frac{\sqrt{7(a+h)}-\sqrt{7a}}{h} $$ I have tried to solve this by multiplying the square roots to both sides but i simply can't solve it and i would appreciate some help and guidance .

Answer 1

Hint: You may always multiply by one or add zero and it will not change anything. Here, let us multiply by $1$ in the form $$\frac{\sqrt{7(a+h)}+\sqrt{7a}}{\sqrt{7(a+h)}+\sqrt{7a}}$$

What happens?

$$\lim\limits_{h\to 0}\frac{\sqrt{7(a+h)}-\sqrt{7a}}{h}\frac{\sqrt{7(a+h)}+\sqrt{7a}}{\sqrt{7(a+h)}+\sqrt{7a}} = \lim\limits_{h\to 0}\frac{7a+7h-7a}{h\left(\sqrt{7(a+h)}+\sqrt{7a}\right)}=\dots$$

Answer 2

$$\lim _{h\to 0}\left(\frac{\sqrt{7\left(a+h\right)}-\sqrt{7a}}{h}\right)$$ Apply L'Hopital's Rule:

$$\frac{d}{dh}\left(\sqrt{7\left(a+h\right)}-\sqrt{7a}\right)=\frac{\sqrt{7}}{2\sqrt{a+h}}$$

$$\frac{d}{dh}\left(h\right)=1$$

Then $$\lim _{h\to 0}\left(\frac{\frac{\sqrt{7}}{2\sqrt{a+h}}}{1}\right) = \lim _{h\to 0}\left(\frac{\sqrt{7}}{2\sqrt{a+h}}\right)= \frac{\sqrt{7}}{2\sqrt{a}}$$