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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.

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  • $\begingroup$ Please write down your limit more precisely. $\endgroup$
    – MrYouMath
    Commented Mar 7, 2017 at 14:12

3 Answers 3

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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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  • $\begingroup$ Got,thanks, I forgot $ {(25 - x^2)}^' = - 2 x $. Is there any way without using derivatives because this question is before derivatives chapter to me. $\endgroup$
    – Noddy
    Commented Mar 7, 2017 at 14:16
  • $\begingroup$ I didn't get 2nd way. Are you rationalizing? You forgot to multiply denominator? (Answer is not =2) $\endgroup$
    – Noddy
    Commented Mar 7, 2017 at 14:23
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$\lim_{x\rightarrow 3}\frac{f(x)-f(1)}{x-1}=\lim_{x\rightarrow 3}\frac{-\sqrt{25-x^2}+\sqrt{24}}{x-1}$=$\frac{-\sqrt{25-3^2}+\sqrt{24}}{3-1}$=$\frac{-4+\sqrt{24}}{2}$=$-2+\sqrt{6}$ which is approximately 0.449

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    $\begingroup$ Sorry for inconvenience,it is x tends to 1 , I have edited $\endgroup$
    – Noddy
    Commented Mar 7, 2017 at 14:26
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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 $$\lim _{x \to 1}\frac{(-\sqrt{25-x^2}+2\sqrt6)(\sqrt{25-x^2}+2\sqrt6)}{(x-1){(\sqrt{25-x^2}+2\sqrt6)}}=\frac{24-25+x^2}{(x-1){(\sqrt{25-x^2}+2\sqrt6)}}=\frac{(x+1)(x-1)}{(\sqrt{25-x^2}+2\sqrt6)(x-1)}=\frac{1+1}{(\sqrt{25-1^2}+2\sqrt6)}= \frac{1}{2\sqrt6}$$

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  • $\begingroup$ math.stackexchange.com/questions/2176009/… $\endgroup$
    – Noddy
    Commented Mar 7, 2017 at 14:28
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    $\begingroup$ Oh .. Ok ..I will change it... I also thought it weird that limit was to be evaluated at 3 $\endgroup$
    – LM2357
    Commented Mar 7, 2017 at 14:29

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