Solving limit without L'Hôpital

Question

I need to solve this limit without L'Hôpital's rule. These questions always seem to have some algebraic trick which I just can't see this time.

$$ \lim_{x\to0} \frac{5-\sqrt{x+25}}{x}$$

Could someone give me a hint as to what I need to do to the fraction to make this work? Thanks!

Answer 1

$$\lim_{x\to0} \frac{5-\sqrt{x+25}}{x}=\lim_{x\to0} \frac{(5-\sqrt{x+25)}(5+\sqrt{x+25})}{x(5+\sqrt{x+25})}=\lim_{x\to0} \frac{25-(x+25)}{x(5+\sqrt{x+25})}=-\frac{1}{10}$$

Answer 2

Definition of derivative at $x=0$ for $f(x) = -\sqrt{x+25}$.

Answer 3

Let $t^2=x+25$, then $t=\sqrt{x+25}.$ Then we have $$\lim_{t\to 5}\frac{5-t}{t^2-25}=\lim_{t\to 5}\dfrac{5-t}{(t+5)(t-5)}=-\lim_{t\to 5}\dfrac{5-t}{(t+5)(5-t)}=-\lim_{t\to 5}\dfrac{1}{t+5}=-\dfrac{1}{10}.$$

Answer 4

Another possible method is the expand the square root term by Taylor Series (also known as Taylor expansion) after taking 25 out of the sqrt. Have you covered Taylor expansion? It's a life saver in many situations!!

Answer 5

$$\lim_{x\to0}\frac{5-\sqrt{x+25}}{x}=\lim_{x\to0}\frac{5-\sqrt{x+25}}{(\sqrt{x+25})²-25}=\lim_{x\to0}\frac{5-\sqrt{x+25}}{-(5-\sqrt{x+25})(5+\sqrt{x+25})}=\lim_{x\to0}\frac{-1}{5+\sqrt{x+25}}=\frac{-1}{10}$$