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$$\lim _{x\to \frac{\pi }{4}}\left(\frac{\cos ^3(x)-\sin ^3(x)}{\sin \left(\frac{\pi }{4}-x\right)}\right)$$

I tried with it for a while and I got minus infinity. Yet, when I put it on wolframalpha, I got $\frac{3}{\sqrt{2}}$ I do not understand where I did the mistake.

Note: Please do not use L'Hopital, it's not allowed for us. Thanks in advance.

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2 Answers 2

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You must use some identities.

$$\left\{ \matrix{ {\sin ^3}(x) - {\cos ^3}(x) = \left( {\sin (x) - \cos (x)} \right)\left( {{{\sin }^2}(x) + \sin (x)\cos (x) + {{\cos }^2}(x)} \right) \hfill \cr \sin (x - {\pi \over 4}) = \sin (x)\cos ({\pi \over 4}) - \cos (x)\sin ({\pi \over 4}) = {{\sqrt 2 } \over 2}\left( {\sin (x) - \cos (x)} \right) \hfill \cr} \right.$$

and hence you have

$$\eqalign{ & \mathop {\lim }\limits_{x \to {\pi \over 4}} {{{{\sin }^3}(x) - {{\cos }^3}(x)} \over {\sin (x - {\pi \over 4})}} = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{\left( {{{\sin }^2}(x) + \sin (x)\cos (x) + {{\cos }^2}(x)} \right)\left( {\sin (x) - \cos (x)} \right)} \over {{{\sqrt 2 } \over 2}\left( {\sin (x) - \cos (x)} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{\left( {{{\sin }^2}(x) + \sin (x)\cos (x) + {{\cos }^2}(x)} \right)} \over {{{\sqrt 2 } \over 2}}} = {2 \over {\sqrt 2 }}\left( {1 + {1 \over 2}} \right) = {3 \over {\sqrt 2 }} \cr} $$

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    $\begingroup$ It's backwards from the problem... but still the right idea $\endgroup$
    – Peter Woolfitt
    Oct 5, 2015 at 19:46
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    $\begingroup$ @H. R. Wow, many thanks. Haha I feel like I'm so stupid in this community :v there are amazing guys here. I love this community ! $\endgroup$
    – Amine Marzouki
    Oct 5, 2015 at 19:50
  • $\begingroup$ @Amine Marzouki: You are welcomed! :) Would you please check-mark my answer. Note the check-mark next to the answer. :) $\endgroup$
    – Hosein Rahnama
    Oct 5, 2015 at 19:53
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    $\begingroup$ @H.R. I did it, thanks once again $\endgroup$
    – Amine Marzouki
    Oct 5, 2015 at 19:57
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    $\begingroup$ @MarkWatson Yes yes, of course. $\endgroup$
    – Amine Marzouki
    Oct 5, 2015 at 20:13
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This one isn't too bad if you recall the trigonometic formula $$\sin(y-x)=\sin(y)\cos(x)-\sin(x)\cos(y)$$ Now we just apply this to the denominator, getting $$\sin\left(\frac{\pi}{4}-x\right)=\frac{1}{\sqrt{2}}\left(\cos(x)-\sin(x)\right)$$

Now to make matters convenient, let's use the substitution $a:=\cos(x)$ and $b:=\sin(x)$.

Your fraction then becomes $$\frac{\cos^3(x)-\sin^3(x)}{\sin\left(\frac{\pi}{4}-x\right)}=\frac{a^3-b^3}{\frac{1}{\sqrt{2}}\left(a-b\right)}=\frac{(a-b)\left(a^2+ab+b^2\right)}{\frac{1}{\sqrt{2}}\left(a-b\right)}=\sqrt{2}\left(a^2+ab+b^2\right)$$

We can directly plug in values to this equation. When $x=\frac{\pi}{4}$, we have $a=b=\frac{1}{\sqrt{2}}$, giving the result $\frac{3}{\sqrt{2}}$.

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  • $\begingroup$ Thank you so much, it's the same method like H.R. ! $\endgroup$
    – Amine Marzouki
    Oct 5, 2015 at 19:56
  • $\begingroup$ @AmineMarzouki Indeed :) (I was just most of the way typing it up already and didn't want to stop) $\endgroup$
    – Peter Woolfitt
    Oct 5, 2015 at 19:57
  • $\begingroup$ We need more awesome people like you :) Thank you very much brother. Can you give me some advices, I want to be good at limits, but I don't know an effective method $\endgroup$
    – Amine Marzouki
    Oct 5, 2015 at 19:59
  • $\begingroup$ @AmineMarzouki Well you could keep in mind that a lot of these sorts of limits without lhopital problems rely on algebraic tricks to get rid of the denominator. $\endgroup$
    – Peter Woolfitt
    Oct 5, 2015 at 20:05
  • $\begingroup$ Woolflitt Thanks ! I'll consider it in my future exercises $\endgroup$
    – Amine Marzouki
    Oct 5, 2015 at 20:11

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