How do I solve this limit without l"Hopital?

Question

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

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.

Answer 1

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

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

Answer 2

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 $\frac{3}{\sqrt{2}}$.