<s>[INST] How to find limit of $\lim_{x\to 1} \frac{f(x) - f(1)}{x-1}$ if $f(x)=-\sqrt{25-x^2}$ I have a  question , if  then find  $$\lim_{x\to 1} \frac{f(x) - f(1)}{x-1}$$
I got $f(1)=-\sqrt {24}$. Should I get limit = $\frac{\sqrt{24} - \sqrt{25-x^2}}{x-1}$ but answer is $\frac{1}{\sqrt{24}}$ . Should I use $f'(x)$? If I use $f'(x)$ then I got $-\frac{1}{\sqrt{24}}$ .I should not get minus in solution. 
Please help, Thanks in advance. 
 [/INST] Method $1$. One may recall that, for any differentiable function $f$ near $a$, one has
$$
\lim_{x \to a}\frac{f(x)-f(a)}{x-a}=f'(a),
$$ 
Here we have
$$
f(x)=-\sqrt{25-x^2},\qquad f'(x)=\frac{x}{\sqrt{25-x^2}},\qquad a=1.
$$
Can you finish it?
Method $2$. (without derivatives) One has, a $x \to 1$,
$$
\begin{align}
 \frac{f(x) - f(1)}{x-1}&=\frac{\sqrt{25-1}-\sqrt{25-x^2}}{x-1}
\\\\&=\frac{(\sqrt{25-1}-\sqrt{25-x^2})(\sqrt{25-1}+\sqrt{25-x^2})}{(x-1)(\sqrt{25-1}+\sqrt{25-x^2})}
\\\\&=\frac{((25-1)-(25-x^2))}{(x-1)(\sqrt{25-1}+\sqrt{25-x^2})}
\\\\&=\frac{x^2-1}{(x-1)(\sqrt{25-1}+\sqrt{25-x^2})}.
\end{align}
$$Can you finish it?
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