Selidikilah apakah benar bahwa lim x-> pi/2 sin(pi/4 - x)

Salah, hasilnya adalah $\frac{\sqrt{2}}{2}$

Pembahasan

Untuk menyelidiki apakah benar bahwa $lim_{x \to \pi/2} \sin(\frac{\pi}{4} - x) \tan(x + \frac{\pi}{4}) = 1$, kita akan menggunakan substitusi dan identitas trigonometri.

Langkah 1: Substitusi langsung.
Jika kita substitusi $x = \frac{\pi}{2}$ ke dalam fungsi, kita mendapatkan:
$$ \sin(\frac{\pi}{4} - \frac{\pi}{2}) \tan(\frac{\pi}{2} + \frac{\pi}{4}) = \sin(-\frac{\pi}{4}) \tan(\frac{3\pi}{4}) $$ $$ = (-\frac{\sqrt{2}}{2})(-1) = \frac{\sqrt{2}}{2} $$

Hasil substitusi langsung tidak sama dengan 1. Mari kita coba manipulasi bentuk.

Langkah 2: Gunakan identitas trigonometri.
Kita tahu bahwa $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$. Jadi, $\tan(x + \frac{\pi}{4}) = \frac{\sin(x + \frac{\pi}{4})}{\cos(x + \frac{\pi}{4})}$.
Limit menjadi:
$$ \lim_{x \to \pi/2} \sin(\frac{\pi}{4} - x) \frac{\sin(x + \frac{\pi}{4})}{\cos(x + \frac{\pi}{4})} $$

Mari kita ubah sudutnya:
$ \sin(\frac{\pi}{4} - x) = -\sin(x - \frac{\pi}{4}) $
$ \cos(x + \frac{\pi}{4}) = \cos(\frac{\pi}{2} - (\frac{\pi}{4} - x)) = \sin(\frac{\pi}{4} - x) = -\sin(x - \frac{\pi}{4}) $

Jika $x \to \frac{\pi}{2}$, maka $x - \frac{\pi}{4} \to \frac{\pi}{4}$.

Kita juga tahu $\sin(A+B)$ dan $\cos(A+B)$.
$ \sin(x + \frac{\pi}{4}) = \sin x \cos \frac{\pi}{4} + \cos x \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}(\sin x + \cos x) $
$ \cos(x + \frac{\pi}{4}) = \cos x \cos \frac{\pi}{4} - \sin x \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}(\cos x - \sin x) $

Dan $ \sin(\frac{\pi}{4} - x) = \sin \frac{\pi}{4} \cos x - \cos \frac{\pi}{4} \sin x = \frac{\sqrt{2}}{2}(\cos x - \sin x) $.

Substitusikan kembali ke dalam limit:
$$ \lim_{x \to \pi/2} \frac{\sqrt{2}}{2}(\cos x - \sin x) \frac{\frac{\sqrt{2}}{2}(\sin x + \cos x)}{\frac{\sqrt{2}}{2}(\cos x - \sin x)} $$ $$ = \lim_{x \to \pi/2} \frac{\sqrt{2}}{2}(\sin x + \cos x) $$

Sekarang substitusi $x = \frac{\pi}{2}$:
$$ \frac{\sqrt{2}}{2}(\sin \frac{\pi}{2} + \cos \frac{\pi}{2}) = \frac{\sqrt{2}}{2}(1 + 0) = \frac{\sqrt{2}}{2} $$