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. 
 A: 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?
A: $\lim_{x\rightarrow 3}\frac{f(x)-f(1)}{x-1}=\lim_{x\rightarrow 3}\frac{-\sqrt{25-x^2}+\sqrt{24}}{x-1}$=$\frac{-\sqrt{25-3^2}+\sqrt{24}}{3-1}$=$\frac{-4+\sqrt{24}}{2}$=$-2+\sqrt{6}$ which is approximately 0.449
A: First, let us evaluate $f(1)=f(x)=-\sqrt{25-1^2}=-2\sqrt{6}$
So we wish to evaluate, 
$$\lim _{x \to 3}\frac{-\sqrt{25-x^2}+2\sqrt6}{x-1}=\frac{-\sqrt{25-3^2}+2\sqrt6}{3-1}=\frac{-4+2\sqrt6}{2}= -2 + \sqrt{6}$$
$EDIT$
$$\lim _{x \to 1}\frac{-\sqrt{25-x^2}+2\sqrt6}{x-1}$$
Rationalizing 
$$\lim _{x \to 1}\frac{(-\sqrt{25-x^2}+2\sqrt6)(\sqrt{25-x^2}+2\sqrt6)}{(x-1){(\sqrt{25-x^2}+2\sqrt6)}}=\frac{24-25+x^2}{(x-1){(\sqrt{25-x^2}+2\sqrt6)}}=\frac{(x+1)(x-1)}{(\sqrt{25-x^2}+2\sqrt6)(x-1)}=\frac{1+1}{(\sqrt{25-1^2}+2\sqrt6)}=
\frac{1}{2\sqrt6}$$
