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We know that the derivative $f'(1)=3$.

$$ \lim_{h \to 0} \frac{f(1-5h^2)-f(1+3h^2)}{h^2(h+1)}=? $$

I try to solve it by applying L'Hôpital's rule, but answer was incorrect.

Since $f$ is differentiable at $x=1$, it has Continuity, so its right and left limit at $x=1$ are equal. So the numerator and denominator are zero. I think we can use L'Hôpital's rule.

With given $f'(1)=3$ $$ \lim_{h \to 0} \frac{f(1-5h^2)-f(1+3h^2)}{h^2(h+1)}=\frac {0}{0}, $$

$$ \lim_{h \to 0} \frac{-10hf(1-5h^2)-6hf(1+3h^2)}{3h^2+1}=\frac {-16 h f'(1)}{3 h ^ 2 +1}=\frac {0}{1}=0, $$ $$ {h \to 0} \implies f'(1-5h^2)=f'(1+3h^2). $$

What is wrong with this answer?

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  • $\begingroup$ Do you know Taylor series ? If yes, did you try ? $\endgroup$ – Claude Leibovici Jan 3 '15 at 8:09
  • $\begingroup$ know I dont know. I just use hopital, but my answer was zero. $\endgroup$ – user123 Jan 3 '15 at 8:21
  • $\begingroup$ L'Hopital only applies if the limit is of the form $0/0$ or $\infty/\infty$, otherwise it is not valid. Since we don't know anything about $f$, you can't blindly apply L'Hopital without more information. $\endgroup$ – Ruvi Lecamwasam Jan 3 '15 at 8:28
  • $\begingroup$ @Andrew Isn't the limit 0/0 right now? $\endgroup$ – user25004 Jan 3 '15 at 8:39
  • $\begingroup$ Please do type it out, don't put the picture instead of equations. $\endgroup$ – Ruslan Jan 3 '15 at 15:20
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$$E=\lim_{h\to 0}\dfrac{f(1-5h^2)-f(1+3h^2)}{h^3+h^2}$$ $$E=\lim_{h\to 0}\dfrac{[f(1-5h^2)-f(1)]-[f(1+3h^2)-f(1)]}{h^3+h^2}$$ $$E=\Big(\lim_{h\to 0}\dfrac{f(1-5h^2)-f(1)}{5h^2}.\dfrac{5h^2}{h^3+h^2}\Big)-\Big(\lim_{h\to 0}\dfrac{f(1+3h^2)-f(1)}{3h^2}.\dfrac{3h^2}{h^3+h^2}\Big)$$ $$E=-5f'(1)-3f'(1)$$

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  • $\begingroup$ The $+$ in the middle should be a $-$, I guess. Fix the typo and the result. Nice answer ! Cheers. $\endgroup$ – Claude Leibovici Jan 3 '15 at 8:23
  • $\begingroup$ I add my answer $\endgroup$ – user123 Jan 3 '15 at 8:56
  • $\begingroup$ @user123: Your mistake is $$-10hf'(1-5h^2)-6hf'(1+3h^2)=-16hf'(1).$$ $\endgroup$ – Bumblebee Jan 3 '15 at 9:51
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Hint: $$ \lim_{h \to 0} \frac{f(1+3h^2)- f(1-5h^2)}{8h^2}= f'(1)$$

Can you substitute this inside?

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  • $\begingroup$ where dose 8h^2 come from? $\endgroup$ – user123 Jan 3 '15 at 8:23
  • $\begingroup$ That is the difference between $-5h^2$ and $3h^2$. You can do it in 2 steps as suggested by Nilan. Although the sign there is currently wrong. $\endgroup$ – user25004 Jan 3 '15 at 8:29
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In the third line of your notes, the denominator is supposed to be the derivative of $h^2(h+1)$; this is $3h^2+2h$ and not $3h^2+1$ as you wrote. If you fix this, you get the result.

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  • $\begingroup$ oh, thanks. I made a mistake. $\endgroup$ – user123 Jan 3 '15 at 9:59
  • $\begingroup$ If I had received a dollar for each of my mistakes, I should be a billionaire !! $\endgroup$ – Claude Leibovici Jan 3 '15 at 10:00
  • $\begingroup$ @OK :) I solve it many times but didn't notice that mistake :) $\endgroup$ – user123 Jan 3 '15 at 10:19

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