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
$$\lim_{x\to 0}\frac{25-(x+25)}{x(5+\sqrt{x+25})}=\lim_{x\to 0}\frac{-x}{x(5+\sqrt{x+25})}=\lim_{x\to 0}\frac{-1}{5+\sqrt{x+25}}$$
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 $x=0$.
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