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Calculus · real student question

Evaluate the limit of (sin 4x + cos 2x)/(4x - pi) as x approaches pi/4.

Question

Evaluate

lim⁡x→π4sin⁡4x+cos⁡2x4x−π\lim_{x\to \frac{\pi}{4}}\frac{\sin 4x+\cos 2x}{4x-\pi}

Step-by-step solution

  1. Check the form. At x=π4x=\frac{\pi}{4} we get sin⁡π+cos⁡π2=0+0=0\sin\pi+\cos\frac{\pi}{2}=0+0=0 on top and π−π=0\pi-\pi=0 underneath, so the form is 00\frac{0}{0}.

  2. Shift the variable to the origin. Put x=π4+tx=\frac{\pi}{4}+t with t→0t\to 0. The denominator becomes 4t4t, which is the simplest possible form to divide by.

  3. Rewrite the numerator. sin⁡4x=sin⁡(π+4t)=−sin⁡4t\sin 4x=\sin(\pi+4t)=-\sin 4t and cos⁡2x=cos⁡(π2+2t)=−sin⁡2t\cos 2x=\cos\left(\frac{\pi}{2}+2t\right)=-\sin 2t, so the numerator is −sin⁡4t−sin⁡2t-\sin 4t-\sin 2t.

  4. Divide by tt. The quotient equals −sin⁡4tt−sin⁡2tt4\frac{-\frac{\sin 4t}{t}-\frac{\sin 2t}{t}}{4}, and by the standard limit sin⁡ktt→k\frac{\sin kt}{t}\to k the numerator tends to −4−2=−6-4-2=-6.

  5. Finish. The limit is −64=−32\frac{-6}{4}=-\frac{3}{2}. Differentiating top and bottom once (L'Hopital) gives 4cos⁡4x−2sin⁡2x4→−4−24\frac{4\cos 4x-2\sin 2x}{4}\to\frac{-4-2}{4}, the same value.

  6. Numerical check. At x=π4+10−7x=\frac{\pi}{4}+10^{-7} the original expression evaluates to −1.5000000-1.5000000, confirming −32-\frac{3}{2}.

Answer

−32-\frac{3}{2}

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