Solving limit without L'Hôpital

Solution 1:

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

Solution 2:

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

Solution 3:

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

Solution 4:

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