Aku n MATEMATIKA cah SMA Nda: Ku Suka LIMIT 3gonome3

Pernah aq baca di semacam kisi-kisi masuk PT di Singapore

tentang limit trigonometri....isinya ringkas banget yaitu :

$sin\, \square \approx \square \: \: \: and\: \: \: tg\, \square \approx \square\: \: untuk\: \: \square \approx 0$
Bagaimanapun juga dalam menyelesaikan persoalan
limit trigono ini qt wajib mengerti ttg identitas- identitasnya
dan konsep dasar menyelesaikan limit trigono tersebut

okleh lah qt pake adja untuk menyelesaikan soal - soal berikut :

$\ast )\: \: \lim_{x\rightarrow \frac{\pi }{2}}\frac{4.(x-\pi ).cos^{2}x}{\pi .(\pi -2x).tg(x-\frac{\pi }{2})}=....$
Jika qt masukkan langsung batasnya maka qt jumpai bentuk Tak-Tentu (T.T)
$\fn_cm sifat\, pada\, kuadran\, I :\, cos\, x=sin(\frac{\pi}{2}-x)$
$\fn_cm \lim_{x\rightarrow \frac{\pi }{2}}\frac{4.(x-\pi ).sin(\frac{\pi}{2}-x).sin(\frac{\pi}{2}-x)}{\pi .(\pi -2x).tg(x-\frac{\pi }{2})}$
$\fn_cm \lim_{x\rightarrow \frac{\pi }{2}}\frac{4.(x-\pi ).(\frac{\pi}{2}-x).(\frac{\pi}{2}-x)}{2.\pi .(\frac{\pi}{2} -x).(x-\frac{\pi }{2})}$
$\fn_cm \frac{4.(\frac{\pi }{2}-\pi ).(-1)}{2.\pi }=\, 1$

$\fn_cm \ast )\: \lim_{x\rightarrow 0}\left \{ \frac{1}{x.tg\, x}-\frac{cos^{2}x}{x.sin\, x} \right \}=....$
Jika qt masukkan langsung batasnya maka qt jumpai bentuk Tak-Tentu (T.T)
coba kita ingat lagi apa yang qt dapat di kelas XI tentang sudut rangkap dua
$\fn_cm cos\, 2x=1-2sin^{2}x\: \: jika \: qt\: tulis\: dalam\: bentuk\: umum$
$\fn_cm cos\, \square =1-2sin^{2}\frac{\square }{2}\: \: atau \: dapat \: qt \: tulis\:\: 2.sin\frac{\square }{2}.sin\frac{\square }{2}=1- cos\, \square$
$\fn_cm untuk\:\square \approx 0\: maka\: 2.sin\frac{\square }{2}.sin\frac{\square }{2}=1- cos\, \square\: dapat \: ditulis$
$\fn_cm 2.\frac{\square }{2}.\frac{\square }{2}={\color{Red} 1- cos\, \square=\frac{\square ^{2}}{2}}$

Mari qt kembali ke soal :
$\fn_cm {\color{Red} INGAT}\: \: sin^{2}\blacksquare +cos^{2}\blacksquare=1$
$\fn_cm \: \frac{sin\, \blacksquare }{cos\, \blacksquare}=tg\, \blacksquare\: \: dan\: \: \frac{cos\, \blacksquare }{sin\, \blacksquare}=cotg\, \blacksquare=\frac{1 }{tg\, \blacksquare}$
$\dpi{100} \fn_cm \lim_{x\rightarrow 0}\left \{ \frac{1}{x.tg\, x}-\frac{cos\, x}{x.tg\, x} \right \}=\lim_{x\rightarrow 0}\frac{1}{x.tg\, x}.(1-cos\, x)$
$\dpi{100} \fn_cm =\frac{1}{x. x}.(\tfrac{1}{2})x^{2}=\frac{1}{2}$

$\fn_cm \ast )\: \: \lim_{x\rightarrow 0}\frac{1-cos^{2}x-cos\, x.sin^{2}x}{x^{4}}=....$
jawab :
$\fn_cm \lim_{x\rightarrow 0}\frac{sin^{2}x-cos\, x.sin^{2}x}{x^{4}}=\lim_{x\rightarrow 0}\frac{sin^{2}x.(1-cos\, x)}{x^{4}}=$
$\fn_cm \lim_{x\rightarrow 0}\frac{x^{2}.(\tfrac{1}{2}x^{2})}{x^{4}}=\frac{1}{2}$

$\fn_cm \ast )\: \: \lim_{t\rightarrow 2}\: \: \frac{(t^{2}-5t+6).sin(t-2)}{(t^{2}-t-2)^{2}}=....$
Jawab : Jika qt masukkan langsung batasnya maka qt jumpai bentuk Tak-Tentu (T.T)
$\fn_cm \lim_{t\rightarrow 2}\: \: \frac{(t-2).(t-3).(t-2)}{(t-2)^{2}.(t+1)^{2}}=\lim_{t\rightarrow 2}\: \: \frac{(t-3)}{(t+1)^{2}}=-\frac{1}{9}$

$\fn_cm \ast )\: \: \lim_{x\rightarrow 2}\: \: \frac{1-cos(x+2)}{x^{2}+4x+4}=....$
Jawab : Jika qt masukkan langsung batasnya maka qt jumpai bentuk Tak-Tentu (T.T)
$\fn_cm \ast )\: \: \lim_{x\rightarrow 2}\: \: \frac{\frac{1}{2}.(x+2)^{2}}{(x+2)^{2}}=\frac{1}{2}$

$\fn_cm \ast )\: \: \lim_{x\rightarrow a}\frac{sin\, x-sin\, a}{x-a}=....$
bentuk diatas bisa dikerjakan dengan memakai definisi sbb :
$\fn_cm \lim_{t\rightarrow p}\frac{f(t)-f(p)}{t-p}=f'(p).....f'adalah\: turunan \: pertama \: f$
Bentuk soal diatas f(x) = sin x sehingga f'(x) = cos x
jadi jawabane adalah cos a karena x mendekati a

$\dpi{100} \fn_cm \ast )\: \: \lim_{x\rightarrow \frac{\pi }{2}}\frac{\sqrt{1-cos(\pi -2x)}}{\left | x- \frac{\pi }{2}\right |}=....$
Jawab : Jika qt masukkan langsung batasnya maka qt jumpai bentuk Tak-Tentu (T.T)
$\dpi{150} \fn_cm {\color{Red} INGAT \, !!!}\: \: \left | \blacksquare \right |=\sqrt{\blacksquare ^{2}}\neq \left (\sqrt{\blacksquare } \right )^{2}$
$\dpi{150} \fn_cm cos\, (-\alpha )=cos\, \alpha$
$\dpi{100} \fn_cm jadi\, cos\, (\pi -2x )=cos\, \left \{ -2(x-\frac{\pi }{2}) \right \} =cos\, \left \{ 2(x-\frac{\pi }{2}) \right \}$
$\dpi{100} \fn_cm \lim_{x\rightarrow \frac{\pi }{2}}\frac{\sqrt{\frac{1}{2}\left \{ 2(x-\frac{\pi }{2}) \right \}^{2}}}{\left | x- \frac{\pi }{2}\right |}=\lim_{x\rightarrow \frac{\pi }{2}}\frac{\sqrt{2.\left \{ (x-\frac{\pi }{2}) \right \}^{2}}}{\left | x- \frac{\pi }{2}\right |}$
$\dpi{100} \fn_cm \lim_{x\rightarrow \frac{\pi }{2}}\frac{\sqrt{2}.\sqrt{\left \{ (x-\frac{\pi }{2}) \right \}^{2}}}{\left | x- \frac{\pi }{2}\right |}=\lim_{x\rightarrow \frac{\pi }{2}}\frac{\sqrt{2}.{ \left |(x-\frac{\pi }{2}) \right |}}{\left | x- \frac{\pi }{2}\right |}=\sqrt{2}$

Coba yang ini dijawab
Ayo semangat......

$\dpi{100} \fn_cm \ast )\: \: \lim_{x\rightarrow 3}\frac{(x^{2}-9).(\sqrt{2x^{2}-2x-3}-3)}{(x^{2}-2x-3).sin\, (2x-6)}=....$

$\dpi{100} \fn_cm \ast )\: \: \lim_{x\rightarrow 0}\: \frac{cos\, 3x-cos\, 5x}{x^{2}}=....$

$\dpi{100} \fn_cm \ast )\: \: \lim_{x\rightarrow 0}\: \frac{\sqrt{1+sin\, x}-\sqrt{1+tg\, x}}{x^{3}.cos\, 2x}=....$

$\dpi{100} \fn_cm \ast )\: \: \lim_{x\rightarrow 0}\: \frac{x.sin\, 5x}{2-\sqrt{3+cos\, x.cos\, 3x}}=....$

$\dpi{100} \fn_cm \ast )\: \: \lim_{x\rightarrow 0}\: \frac{1-cos^{7}x}{1-\sqrt{cos\, x}}=....$

$\fn_cm \ast )\: \: \lim_{x\rightarrow \frac{\pi }{2}}\frac{sin(x-\frac{\pi }{2})}{\sqrt{\frac{x }{2}}-\sqrt{\frac{\pi }{4}}}=...$

$\fn_cm \ast )\: \: \lim_{x\rightarrow \infty }\: x^{2}.sin(\frac{1}{x}).tg(\frac{1}{x})=....$

$\fn_cm \ast )\: \: \lim_{x\rightarrow 0}\: \frac{sin\, 4x}{\sqrt{1-x}-1}=....$

$\fn_cm \ast )\: \: \lim_{x\rightarrow 1}\: \frac{x^{2}+x-2}{sin\, (2-\sqrt{x+3})}=....$

$\fn_cm \ast )\: \: \lim_{x\rightarrow \pi }\: \frac{\sqrt{5+cosx}-2}{(\pi -x)^{2}}=....$

soal-soal diatas dapat di download disini