How is this limit calculated without l'Hospital?

Question

In this question: Solving limit without L'Hôpital

The answer is as follows:

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

Expanding the fraction makes sense, but I dont understand how we get $ -\frac{1}{10}$ as a result. Because when you put in 0 for x ( which I intuitively did) I get $\frac{0}{0}$ as a result, which doesnt get me anywhere withou l'Hospital.

What step did I miss ?

Answer 1

$$\lim_{x\to 0}\frac{25-(x+25)}{x(5+\sqrt{x+25})}=\lim_{x\to 0}\frac{-x}{x(5+\sqrt{x+25})}=\lim_{x\to 0}\frac{-1}{5+\sqrt{x+25}}$$ Note that after the second step we cancel out the $x$ in the numerator and denominator, and are left with an expression with which we can evaluate at $x=0$.

Answer 2

other If $f (x)=\sqrt {x+25} $ then $$\lim_0\frac {f (0)-f (x)}{x}=-f'(0) $$ and $$f'(x)=\frac {1}{2f (x)} $$

Answer 3

Hint:

Let $5-\sqrt{x+25}=y$

$\implies(i)5-y=\sqrt{x+25}\implies x=y^2-10y$

and $(ii)y\to0^-$

Can you take it from here?