Pembahasan:

\[\lim_{x \to 0} \frac{\sin 3x}{x} = \lim_{x \to 0} \frac{3 \cdot \sin 3x}{3x} = 3 \cdot \lim_{x \to 0} \frac{\sin 3x}{3x} = 3 \cdot 1 = \boxed{3}\]

Pembahasan:

\[\lim_{x \to 0} \frac{\sin 5x}{\sin 2x} = \frac{5}{2} = \boxed{\frac{5}{2}}\]

Menggunakan rumus \(\lim_{x\to 0}\frac{\sin ax}{\sin bx}=\frac{a}{b}\).

Pembahasan:

\[\lim_{x \to 0} \frac{4x}{\sin 2x} = \frac{4}{2} \cdot \lim_{x\to 0}\frac{2x}{\sin 2x} = 2 \cdot 1 = \boxed{2}\]

Pembahasan:

\[\lim_{x \to 0} \frac{\sin x}{3x} = \frac{1}{3}\cdot\lim_{x\to 0}\frac{\sin x}{x} = \frac{1}{3}\cdot 1 = \boxed{\frac{1}{3}}\]

Pembahasan:

\[\lim_{x \to 0} \frac{\sin 4x}{4x} = 1\]

Langsung menggunakan rumus dasar \(\lim_{x\to 0}\frac{\sin u}{u}=1\) dengan \(u=4x\). Jawaban: \(\boxed{1}\)

Pembahasan:

\[\lim_{x \to 0} \frac{\sin 3x – \sin x}{x} = \lim_{x\to 0}\frac{\sin 3x}{x} – \lim_{x\to 0}\frac{\sin x}{x} = 3 – 1 = \boxed{2}\]

Pembahasan:

\[\lim_{x \to 0} \frac{\sin^2 2x}{x^2} = \lim_{x\to 0}\left(\frac{\sin 2x}{x}\right)^2 = \left(\frac{2}{1}\right)^2 = \boxed{4}\]

Pembahasan:

\[\frac{x \sin 2x}{\sin^2 3x} = \frac{x}{\sin 3x}\cdot\frac{\sin 2x}{\sin 3x}\]

\[= \frac{1}{3}\cdot\frac{3x}{\sin 3x}\cdot\frac{2}{3}\cdot\frac{\sin 2x}{2x}\cdot\frac{3x}{\sin 3x}\]

Saat \(x\to 0\): \(= \frac{1}{3}\cdot 1 \cdot \frac{2}{3}\cdot 1 \cdot 1 = \boxed{\frac{2}{9}}\)

Pembahasan:

\[= \lim_{x\to 0}\frac{\sin 5x}{\sin 4x} + \lim_{x\to 0}\frac{\sin 3x}{\sin 4x} = \frac{5}{4}+\frac{3}{4} = \boxed{2}\]

Pembahasan:

\[= \lim_{x\to 0}\frac{\sin 2x}{x}\cdot\lim_{x\to 0}\frac{\sin 3x}{x} = 2 \cdot 3 = \boxed{6}\]

Pembahasan:

Gunakan identitas \(1-\cos 2x = 2\sin^2 x\):

\[\lim_{x\to 0}\frac{2\sin^2 x}{x\sin x} = \lim_{x\to 0}\frac{2\sin x}{x} = 2\cdot 1 = \boxed{2}\]

Pembahasan:

Substitusi \(x = \frac{\pi}{6}+h\), saat \(h\to 0\):

\(2\sin x – 1 = 2\sin(\tfrac{\pi}{6}+h)-1 = 2(\tfrac{1}{2}\cos h + \tfrac{\sqrt{3}}{2}\sin h)-1 = \cos h – 1 + \sqrt{3}\sin h\)

\(\sin 3x = \sin(\tfrac{\pi}{2}+3h) = \cos 3h\)

Saat \(h\to 0\): pembilang \(\to 0 + \sqrt{3}\cdot 0 = 0\) tapi \(\cos 3h \to 1\).

Lebih teliti: \(\frac{\cos h -1+\sqrt{3}\sin h}{\cos 3h}\). Saat \(h\to 0\), penyebut \(\to 1\), jadi kita evaluasi pembilang. \(\cos h – 1 \approx -\frac{h^2}{2}\), \(\sqrt{3}\sin h \approx \sqrt{3}h\).

Pembilang \(\approx \sqrt{3}h\), penyebut \(\to 1\). Jadi limit \(= 0\)?

Periksa: substitusi langsung \(x=\frac{\pi}{6}\): pembilang \(= 2\cdot\frac{1}{2}-1=0\), penyebut \(=\sin\frac{\pi}{2}=1\).

Bukan bentuk \(\frac{0}{0}\), maka substitusi langsung:

\[\frac{2\sin\frac{\pi}{6}-1}{\sin\frac{\pi}{2}} = \frac{0}{1} = \boxed{0}\]

Pembahasan:

Gunakan deret Taylor: \(\sin x = x – \frac{x^3}{6}+\cdots\), \(\cos x = 1-\frac{x^2}{2}+\cdots\)

\[\sin x – x\cos x = \left(x-\frac{x^3}{6}+\cdots\right) – x\left(1-\frac{x^2}{2}+\cdots\right)\]

\[= x – \frac{x^3}{6} – x + \frac{x^3}{2} + \cdots = \frac{x^3}{3}+\cdots\]

\[\lim_{x\to 0}\frac{\frac{x^3}{3}}{x^3} = \boxed{\frac{1}{3}}\]

Pembahasan:

Deret Taylor: \(x – \sin x = x – (x – \frac{x^3}{6}+\cdots) = \frac{x^3}{6}+\cdots\)

\(x^2\sin x = x^2(x – \frac{x^3}{6}+\cdots) = x^3 – \frac{x^5}{6}+\cdots\)

\[\lim_{x\to 0}\frac{\frac{x^3}{6}}{x^3} = \boxed{\frac{1}{6}}\]

Pembahasan:

Misalkan \(u = \sin x\), saat \(x\to 0\) maka \(u\to 0\):

\[\lim_{x\to 0}\frac{\sin(\sin x)}{x} = \lim_{x\to 0}\frac{\sin(\sin x)}{\sin x}\cdot\frac{\sin x}{x}\]

\[= \lim_{u\to 0}\frac{\sin u}{u}\cdot\lim_{x\to 0}\frac{\sin x}{x} = 1 \cdot 1 = \boxed{1}\]

Pembahasan:

\[\frac{1-\cos x}{x^2} = \frac{2\sin^2\frac{x}{2}}{x^2} = \frac{2\sin^2\frac{x}{2}}{4\cdot\frac{x^2}{4}} = \frac{1}{2}\left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2\]

Saat \(x\to 0\): \(= \frac{1}{2}\cdot 1^2 = \boxed{\frac{1}{2}}\)

Pembahasan:

\[\frac{1-\cos x}{x} = \frac{2\sin^2\frac{x}{2}}{x} = \sin\frac{x}{2}\cdot\frac{\sin\frac{x}{2}}{\frac{x}{2}}\cdot\frac{1}{1}\]

Saat \(x\to 0\): \(\sin\frac{x}{2}\to 0\) dan \(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\to 1\), jadi hasilnya \(= 0\cdot 1 = \boxed{0}\)

Pembahasan:

Menggunakan rumus \(\lim_{x\to 0}\frac{1-\cos ax}{x^2}=\frac{a^2}{2}\) dengan \(a=4\):

\[= \frac{4^2}{2} = \frac{16}{2} = \boxed{8}\]

Pembahasan:

\[= \frac{\frac{1-\cos 2x}{x^2}}{\frac{1-\cos 3x}{x^2}} = \frac{\frac{4}{2}}{\frac{9}{2}} = \frac{4}{9} = \boxed{\frac{4}{9}}\]

Pembahasan:

\[= \frac{1-\cos 6x}{x^2}\cdot\frac{x}{\sin 3x}\cdot\frac{x}{1}\cdot\frac{1}{1}\]

Koreksi: \(\frac{1-\cos 6x}{\sin 3x \cdot x} = \frac{1-\cos 6x}{x^2}\cdot\frac{x}{\sin 3x}\)

\(= \frac{36}{2}\cdot\frac{1}{3} = 18\cdot\frac{1}{3} = \boxed{6}\)

Pembahasan:

\(1-\cos^2 x = \sin^2 x\), maka:

\[\lim_{x\to 0}\frac{\sin^2 x}{x^2} = \left(\lim_{x\to 0}\frac{\sin x}{x}\right)^2 = 1^2 = \boxed{1}\]

Pembahasan:

\[\cos 2x – \cos 4x = (1-(1-\cos 2x)) – (1-(1-\cos 4x))\]

\[= -(1-\cos 2x)+(1-\cos 4x) = (1-\cos 4x)-(1-\cos 2x)\]

\[\frac{(1-\cos 4x)-(1-\cos 2x)}{x^2} = \frac{16}{2}-\frac{4}{2} = 8-2 = \boxed{6}\]

Pembahasan:

\(1-\cos 2x = 2\sin^2 x\), maka:

\[\frac{2\sin^2 x}{\sin^2 x} = \boxed{2}\]

Pembahasan:

\[= \frac{1-\cos x}{x^2}\cdot\frac{x^2}{\sin^2 2x} = \frac{1}{2}\cdot\frac{1}{4} = \boxed{\frac{1}{8}}\]

Karena \(\frac{x^2}{\sin^2 2x}=\left(\frac{x}{\sin 2x}\right)^2=\left(\frac{1}{2}\right)^2=\frac{1}{4}\)

Pembahasan:

\(1-\cos 2x = 2\sin^2 x\):

\[\frac{x\cdot 2\sin^2 x}{\sin^3 x} = \frac{2x}{\sin x} = 2\cdot\frac{x}{\sin x}\]

Saat \(x\to 0\): \(= 2\cdot 1 = \boxed{2}\)

Pembahasan:

Gunakan rumus jumlah ke kali: \(\cos A – \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2}\)

\[\cos x – \cos 3x = -2\sin 2x\sin(-x) = 2\sin 2x\sin x\]

\[\frac{2\sin 2x\sin x}{x\sin x} = \frac{2\sin 2x}{x} = 2\cdot\frac{\sin 2x}{x} = 2\cdot 2 = \boxed{4}\]

Pembahasan:

Misalkan \(h = x – \frac{\pi}{2}\), saat \(x\to\frac{\pi}{2}\) maka \(h\to 0\):

\(\cos x = \cos(\frac{\pi}{2}+h) = -\sin h\)

\[\lim_{h\to 0}\frac{-\sin h}{h} = -1 = \boxed{-1}\]

Pembahasan:

Tambah dan kurangi:

\[1-\cos x\cos 2x = (1-\cos x)+\cos x(1-\cos 2x)\]

\[\frac{(1-\cos x)}{x^2}+\frac{\cos x(1-\cos 2x)}{x^2}\]

Saat \(x\to 0\): \(= \frac{1}{2}+1\cdot\frac{4}{2} = \frac{1}{2}+2 = \boxed{\frac{5}{2}}\)

Pembahasan:

\(1+\cos 2x = 2\cos^2 x\), maka \(\sqrt{1+\cos 2x}=\sqrt{2}|\cos x|=\sqrt{2}\cos x\) (dekat 0).

\[\frac{\sqrt{2}\cos x – \sqrt{2}}{x^2} = \sqrt{2}\cdot\frac{\cos x – 1}{x^2} = \sqrt{2}\cdot\left(-\frac{1}{2}\right) = \boxed{-\frac{\sqrt{2}}{2}}\]

Pembahasan:

Tuliskan: \(1-\cos x\cos 2x\cos 3x\)

\(= (1-\cos x)+\cos x(1-\cos 2x)+\cos x\cos 2x(1-\cos 3x)\)

Bagi \(x^2\) dan ambil limit:

\[= \frac{1}{2}+1\cdot\frac{4}{2}+1\cdot 1\cdot\frac{9}{2} = \frac{1}{2}+2+\frac{9}{2} = \frac{1+4+9}{2} = \boxed{7}\]

Pembahasan:

\[\lim_{x\to 0}\frac{\tan 3x}{x} = 3\cdot\lim_{x\to 0}\frac{\tan 3x}{3x} = 3\cdot 1 = \boxed{3}\]

Pembahasan:

\[\lim_{x\to 0}\frac{\tan 2x}{\tan 5x} = \frac{2}{5} = \boxed{\frac{2}{5}}\]

Pembahasan:

\[\lim_{x\to 0}\frac{\sin 3x}{\tan 2x} = \frac{3}{2} = \boxed{\frac{3}{2}}\]

Pembahasan:

\[\frac{6x}{\tan 3x} = \frac{6}{3}\cdot\frac{3x}{\tan 3x} = 2\cdot 1 = \boxed{2}\]

Pembahasan:

\[\frac{\tan 4x}{\sin 4x} = \frac{\sin 4x}{\cos 4x \cdot \sin 4x} = \frac{1}{\cos 4x}\]

Saat \(x\to 0\): \(\frac{1}{\cos 0} = \boxed{1}\)

Pembahasan:

\[\tan 2x – \sin 2x = \frac{\sin 2x}{\cos 2x}-\sin 2x = \sin 2x\left(\frac{1}{\cos 2x}-1\right) = \sin 2x\cdot\frac{1-\cos 2x}{\cos 2x}\]

\[= \frac{\sin 2x \cdot 2\sin^2 x}{\cos 2x}\]

\[\frac{\sin 2x \cdot 2\sin^2 x}{x^3\cos 2x} = \frac{\sin 2x}{x}\cdot\frac{2\sin^2 x}{x^2}\cdot\frac{1}{\cos 2x} = 2\cdot 2\cdot 1 = \boxed{4}\]

Pembahasan:

\[= \frac{\tan 3x}{x}\cdot\frac{\tan 3x}{\sin 2x} = 3\cdot\frac{3}{2} = \boxed{\frac{9}{2}}\]

Pembahasan:

\(1-\cos 2x = 2\sin^2 x\):

\[\frac{2\sin^2 x}{\tan^2 x} = \frac{2\sin^2 x}{\frac{\sin^2 x}{\cos^2 x}} = 2\cos^2 x\]

Saat \(x\to 0\): \(= 2\cdot 1 = \boxed{2}\)

Pembahasan:

\[= \frac{\sin 4x}{3x}+\frac{\tan 2x}{3x} = \frac{4}{3}+\frac{2}{3} = \boxed{2}\]

Pembahasan:

\(1-\cos x = 2\sin^2\frac{x}{2}\):

\[\frac{x\tan x}{2\sin^2\frac{x}{2}} = \frac{x}{\sin\frac{x}{2}}\cdot\frac{\tan x}{\sin\frac{x}{2}}\cdot\frac{1}{2}\]

Alternatif: \(\frac{x\tan x}{1-\cos x} = \frac{x}{1}\cdot\frac{\tan x}{x}\cdot\frac{x^2}{1-\cos x}\cdot\frac{1}{x}\)

Lebih simpel: \(= \frac{x\tan x}{x^2}\cdot\frac{x^2}{1-\cos x} = \frac{\tan x}{x}\cdot\frac{1}{\frac{1-\cos x}{x^2}} = \frac{1}{\frac{1}{2}} = \boxed{2}\)

Pembahasan:

\[\tan x – 1 = \frac{\sin x – \cos x}{\cos x}\]

\[\frac{\tan x – 1}{\sin x – \cos x} = \frac{\sin x – \cos x}{\cos x(\sin x – \cos x)} = \frac{1}{\cos x}\]

Saat \(x\to\frac{\pi}{4}\): \(\frac{1}{\cos\frac{\pi}{4}} = \frac{1}{\frac{\sqrt{2}}{2}} = \boxed{\sqrt{2}}\)

Pembahasan:

\[\tan x – \sin x = \frac{\sin x}{\cos x}-\sin x = \sin x\cdot\frac{1-\cos x}{\cos x}\]

\[\frac{\sin x(1-\cos x)}{x^3\cos x} = \frac{\sin x}{x}\cdot\frac{1-\cos x}{x^2}\cdot\frac{1}{\cos x} = 1\cdot\frac{1}{2}\cdot 1 = \boxed{\frac{1}{2}}\]

Pembahasan:

Ini adalah limit klasik yang terkenal sulit. Menggunakan ekspansi deret Taylor orde tinggi:

\(\sin x = x – \frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+\cdots\)

\(\tan x = x+\frac{x^3}{3}+\frac{2x^5}{15}+\frac{17x^7}{315}+\cdots\)

Setelah perhitungan panjang (substitusi dan ekspansi hingga orde 7):

\[\tan(\sin x)-\sin(\tan x) = -\frac{x^7}{30}+\cdots\]

\[\lim_{x\to 0}\frac{-\frac{x^7}{30}}{x^7} = \boxed{-\frac{1}{30}}\]

Pembahasan:

Faktorkan: \(a^3-b^3 = (a-b)(a^2+ab+b^2)\) dengan \(a=\tan 2x, b=\sin 2x\):

\(\tan 2x – \sin 2x = \sin 2x\cdot\frac{1-\cos 2x}{\cos 2x}\)

Saat \(x\to 0\): \(\tan^2 2x + \tan 2x\sin 2x + \sin^2 2x \approx (2x)^2+(2x)(2x)+(2x)^2 = 12x^2\)

\(\tan 2x – \sin 2x \approx 2x\cdot\frac{2x^2}{1}\cdot 2 = 4x^3\) (lebih tepatnya).

Dengan ekspansi: \(\tan 2x – \sin 2x \approx \frac{(2x)^3}{2}\cdot\frac{1}{1} = 4x^3\)

Maka: \(\frac{4x^3\cdot 12x^2}{x^5} = \frac{48x^5}{x^5} = \boxed{48}\)

Pembahasan:

Substitusi \(x = \frac{\pi}{4}+h\), \(h\to 0\):

\(\tan x = \tan(\frac{\pi}{4}+h) = \frac{1+\tan h}{1-\tan h}\)

\(1-\tan x = 1-\frac{1+\tan h}{1-\tan h} = \frac{-2\tan h}{1-\tan h}\)

\(\sqrt{2}\sin x = \sqrt{2}\sin(\frac{\pi}{4}+h) = \sqrt{2}(\frac{\sqrt{2}}{2}\cos h+\frac{\sqrt{2}}{2}\sin h) = \cos h+\sin h\)

\(1-\sqrt{2}\sin x = 1-\cos h – \sin h\)

Saat \(h\to 0\): pembilang \(\approx \frac{-2h}{1} = -2h\), penyebut \(\approx \frac{h^2}{2}-h = -h+\frac{h^2}{2}\approx -h\)

\[\frac{-2h}{-h} = \boxed{2}\]