Bank Soal Matematika SMA Limit Fungsi Trigonometri

Soal

Pilgan

Nilai limx2(x2)cos(πx2π)tan(4πx8π)=....\lim\limits_{x\rightarrow2}\frac{\left(x-2\right)\cos\left(\pi x-2\pi\right)}{\tan\left(4\pi x-8\pi\right)}=....

A

12π\frac{1}{2\pi}

B

π2\frac{\pi}{2}

C

π\pi

D

14π\frac{1}{4\pi}

E

4π4\pi

Pembahasan:

Subtitusi langsung x=2x=2 menghasilkan bentuk tak tentu 00\frac{0}{0}.

Ingat bahwa

limx0 axtanbx=ab\lim\limits_{x\rightarrow0}\ \frac{ax}{\tan bx}=\frac{a}{b}

Dengan demikian, diperoleh

limx2 (x2)cos(πx2π)tan(4πx8π)=limx2 (x2)cos(πx2π)tan4π(x2)\lim\limits_{x\rightarrow2}\ \frac{\left(x-2\right)\cos\left(\pi x-2\pi\right)}{\tan\left(4\pi x-8\pi\right)}=\lim\limits_{x\rightarrow2}\ \frac{\left(x-2\right)\cos\left(\pi x-2\pi\right)}{\tan4\pi\left(x-2\right)}

                                =limx2 ((x2)tan4π(x2)cosπ(x2))\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\lim\limits_{x\rightarrow2}\ \left(\frac{\left(x-2\right)}{\tan4\pi\left(x-2\right)}\cdot\cos\pi\left(x-2\right)\right)

=limx2 (x2)tan4π(x2)limx2cosπ(x2)=\lim\limits_{x\rightarrow2}\ \frac{\left(x-2\right)}{\tan4\pi\left(x-2\right)}\cdot\lim\limits_{x\rightarrow2}\cos\pi\left(x-2\right)

                                =14π1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{4\pi}\cdot1

                                =14π\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{4\pi}

Jadi, nilai limx2 (x2)cos(πx2π)tan(4πx8π)=14π\lim\limits_{x\rightarrow2}\ \frac{\left(x-2\right)\cos\left(\pi x-2\pi\right)}{\tan\left(4\pi x-8\pi\right)}=\frac{1}{4\pi}


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