||
Membaca
SIMAK UI 2010

Solusi SIMAK UI 2010 Kode 509 [IPA] – Nomor 9


\lim\limits_{x\rightarrow 0}\frac{\sqrt{sin2x+1}-\sqrt{tan2x+1}}{x^3}=....
(A)\quad -8\\(B)\quad -4\\(C)\quad -2\\(D)\quad 2\\(E)\quad 4
Jawaban: C

\begin{array} {lcl} \lim\limits_{x\rightarrow 0}\frac{\sqrt{sin2x+1}-\sqrt{tan2x+1}}{x^3}&=&\lim\limits_{x\rightarrow 0}\frac{(sin2x+1)-(tan2x+1)}{x^3\big(\sqrt{sin2x+1}+\sqrt{tan2x+1}\big)}\\\\&=&\lim\limits_{x\rightarrow 0}\frac{sin2x-tan2x}{x^3\big(\sqrt{sin2x+1}+\sqrt{tan2x+1}\big)}\\\\&=&\lim\limits_{x\rightarrow 0}\frac{sin2x-\frac{sin2x}{cos2x}}{x^3\big(\sqrt{sin2x+1}+\sqrt{tan2x+1}\big)}\\\\&=&\lim\limits_{x\rightarrow 0}\frac{sin2xcos2x-sin2x}{x^3cos2x\big(\sqrt{sin2x+1}+\sqrt{tan2x+1}\big)}\\\\&=&\lim\limits_{x\rightarrow 0}\frac{sin2x(cos2x-1)}{x^3cos2x\big(\sqrt{sin2x+1}+\sqrt{tan2x+1}\big)}\\\\&=&\lim\limits_{x\rightarrow 0}\frac{2sinxcosx(-2sin^{2}x)}{x^3cos2x\big(\sqrt{sin2x+1}+\sqrt{tan2x+1}\big)}\\\\&=&\lim\limits_{x\rightarrow 0}\frac{-4sin^{3}xcosx}{x^3cos2x\big(\sqrt{sin2x+1}+\sqrt{tan2x+1}\big)}\\\\&=&\lim\limits_{x\rightarrow 0}\Big[-4\ .\ \frac{sin^{3}x}{x^3}\ .\ \frac{cosx}{cos2x\big(\sqrt{sin2x+1}+\sqrt{tan2x+1}\big)}\Big]\\\\&=&-4\times (1)^3\times \frac{1}{2}\\\\&=&-2\end{array}

About Kalakay

Guru Matematika SMK

Diskusi

Belum ada komentar.

Tinggalkan Balasan

Isikan data di bawah atau klik salah satu ikon untuk log in:

Logo WordPress.com

You are commenting using your WordPress.com account. Logout / Ubah )

Gambar Twitter

You are commenting using your Twitter account. Logout / Ubah )

Foto Facebook

You are commenting using your Facebook account. Logout / Ubah )

Foto Google+

You are commenting using your Google+ account. Logout / Ubah )

Connecting to %s

%d blogger menyukai ini: