Soal dan Pembahasan Super Lengkap – Limit Menuju Takhingga

Pendekatan formal:
$\begin{array}{rl}& {\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{2x^3+3x^2-5x+4}{2x^4-4x^2+9}}}\\ & =\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{{\displaystyle \frac{2x^3}{x^4}}+{\displaystyle \frac{3x^2}{x^4}}-{\displaystyle \frac{5x}{x^4}}+{\displaystyle \frac{4}{x^4}}}{{\displaystyle \frac{2x^4}{x^4}}-{\displaystyle \frac{4x^2}{x^4}}+{\displaystyle \frac{9}{x^4}}}}\\ & ={\displaystyle \frac{0-0-0+0}{2-0+0}}=0\end{array}$
Pendekatan lain:
Perhatikan bahwa bagian pembilang dan penyebut fungsinya merupakan fungsi polinom. Karena derajat pembilang = $3$ < derajat penyebut = $4,$, nilai limitnya adalah $\boxed{{\displaystyle 0}}$
(Jawaban B)

Pendekatan formal:
Bagi setiap suku dengan variabel berpangkat tertinggi, yaitu $x^3$. 
$$\begin{array}{rl}{\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{2x^3+3x^2+7}{x^2+3x+4}}}& =\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{{\displaystyle \frac{2x^3}{x^3}}+{\displaystyle \frac{3x^2}{x^3}}+{\displaystyle \frac{7}{x^3}}}{{\displaystyle \frac{x^2}{x^3}}+{\displaystyle \frac{3x}{x^3}}+{\displaystyle \frac{4}{x^3}}}}=\mathrm{\infty }\end{array}$$Pendekatan lain:
Perhatikan bahwa bagian pembilang dan penyebut fungsinya merupakan fungsi polinom. Karena derajat pembilang = $3$ > derajat penyebut = $2$, nilai limitnya adalah $\mathrm{\infty }.$
Jadi, nilai dari $\boxed{{\displaystyle {\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{2x^3+3x^2+7}{x^2+3x+4}}=\mathrm{\infty }}}}$
(Jawaban A)

$$\begin{array}{rl}& {\displaystyle \underset{x\to \mathrm{\infty }}{lim}(3-x+{\displaystyle \frac{x^2-2x}{x+5}})}\\ & =\underset{x\to \mathrm{\infty }}{lim}({\displaystyle \frac{(3-x)(x+5)}{x+5}}+{\displaystyle \frac{x^2-2x}{x+5}})\\ & =\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{(3x+15-\cancel{x^2}-5x)+(\cancel{x^2}-2x)}{x+5}}\\ & =\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{-4x+15}{x+5}}={\displaystyle \frac{-4}{1}}=-4\end{array}$$Jadi, nilai dari $\boxed{{\displaystyle {\displaystyle \underset{x\to \mathrm{\infty }}{lim}(3-x+{\displaystyle \frac{x^2-2x}{x+5}})=-4}}}$
(Jawaban A) 

Diketahui bahwa
$\begin{array}{rl}{\displaystyle \frac{f(x)}{x}}& ={\displaystyle \frac{x+{\displaystyle \frac{x^2}{\sqrt{x^2-2x}}}}{x}}\\ & =1+{\displaystyle \frac{x}{\sqrt{x^2-2x}}}.\end{array}$
Dengan demikian, dapat ditulis
$\begin{array}{rl}{\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{f(x)}{x}}}& =\underset{x\to \mathrm{\infty }}{lim}(1+{\displaystyle \frac{x}{\sqrt{x^2-2x}}})\\ & =\underset{x\to \mathrm{\infty }}{lim}(1+{\displaystyle \frac{{\displaystyle \frac{x}{x}}}{\sqrt{{\displaystyle \frac{x^2-2x}{x^2}}}}})\\ & =1+{\displaystyle \frac{1}{\sqrt{1+0}}}=2.\end{array}$
Jadi, nilai dari ${\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{f(x)}{x}}}$ adalah $\boxed{{\displaystyle 2}}$
(Jawaban D)

Bagi setiap suku dengan variabel berpangkat tertinggi, yaitu $x$. 
$$\begin{array}{rl}& {\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{\sqrt{18x^2-x+1}-3x}{\sqrt{x^2+2x}}}}\\ & =\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{{\displaystyle \frac{1}{x}}\sqrt{18x^2-x+1}-{\displaystyle \frac{3x}{x}}}{{\displaystyle \frac{1}{x}}\sqrt{x^2+2x}}}\\ & =\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{{\displaystyle \frac{\sqrt{18x^2-x+1}}{x^2}}-{\displaystyle \frac{3x}{x}}}{{\displaystyle \frac{\sqrt{x^2+2x}}{x^2}}}}\\ & ={\displaystyle \frac{\sqrt{18-0+0}-3}{\sqrt{1+0}}}\\ & ={\displaystyle \frac{\sqrt{18}-3}{1}}\\ & =3\sqrt{2}-3\\ & =3(\sqrt{2}-1)\end{array}$$Jadi, nilai dari $$\boxed{{\displaystyle {\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{\sqrt{18x^2-x+1}-3x}{\sqrt{x^2+2x}}}=3(\sqrt{2}-1)}}}$$(Jawaban B)

Dengan metode pengalian akar sekawan, diperoleh
$$\begin{array}{rl}& {\displaystyle \underset{x\to \mathrm{\infty }}{lim}(\sqrt{x-\sqrt{x}}-\sqrt{x+\sqrt{x}})}\\ & ={\displaystyle \underset{x\to \mathrm{\infty }}{lim}(\sqrt{x-\sqrt{x}}-\sqrt{x+\sqrt{x}})\times {\displaystyle \frac{\sqrt{x-\sqrt{x}}+\sqrt{x+\sqrt{x}}}{\sqrt{x-\sqrt{x}}+\sqrt{x+\sqrt{x}}}}}\\ & =\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{(x-\sqrt{x})-(x+\sqrt{x})}{\sqrt{x-\sqrt{x}}+\sqrt{x+\sqrt{x}}}}\\ & =\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{-2\sqrt{x}}{\sqrt{x-\sqrt{x}}+\sqrt{x+\sqrt{x}}}}.\end{array}$$Bagi setiap sukunya dengan $\sqrt{x}$. 
$${\displaystyle \begin{array}{rl}& \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{{\displaystyle \frac{-2\sqrt{x}}{\sqrt{x}}}}{{\displaystyle \frac{\sqrt{x-\sqrt{x}}}{\sqrt{x}}}+{\displaystyle \frac{\sqrt{x+\sqrt{x}}}{\sqrt{x}}}}}\\ & =\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{-2}{\sqrt{1-{\displaystyle \frac{1}{\sqrt{x}}}}+\sqrt{1+{\displaystyle \frac{1}{\sqrt{x}}}}}}\\ & ={\displaystyle \frac{-2}{\sqrt{1-0}+\sqrt{1+0}}}\\ & ={\displaystyle \frac{-2}{1+1}}=-1\end{array}}$$Jadi, nilai dari $\boxed{{\displaystyle {\displaystyle \underset{x\to \mathrm{\infty }}{lim}(\sqrt{x-\sqrt{x}}-\sqrt{x+\sqrt{x}})=-1}}}$ 
(Jawaban C)

Kalikan dengan bentuk sekawannya, 
$$\begin{array}{rl}{\displaystyle \underset{x\to \mathrm{\infty }}{lim}x(\sqrt{x^2+1}-x)}& =\underset{x\to \mathrm{\infty }}{lim}x(\sqrt{x^2+1}-x)\times {\displaystyle \frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x}}\\ & =\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{x(x^2+1-x^2)}{\sqrt{x^2+1}+x}}\\ & =\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{x}{\sqrt{x^2+1}+x}}\end{array}$$Bagi setiap suku dengan variabel berpangkat tertinggi, yaitu $x$. 
$$\begin{array}{rl}\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{x}{\sqrt{x^2+1}+x}}& =\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{{\displaystyle \frac{x}{x}}}{\sqrt{{\displaystyle \frac{x^2+1}{x^2}}}+{\displaystyle \frac{x}{x}}}}\\ & ={\displaystyle \frac{1}{\sqrt{1+0}+1}}={\displaystyle \frac{1}{2}}\end{array}$$Jadi, nilai dari $\boxed{{\displaystyle {\displaystyle \underset{x\to \mathrm{\infty }}{lim}x(\sqrt{x^2+1}-x)={\displaystyle \frac{1}{2}}}}}$
(Jawaban E)

Dengan menggunakan metode pengalian akar sekawan, diperoleh
$$\begin{array}{rl}& {\displaystyle \underset{x\to \mathrm{\infty }}{lim}(\sqrt{x^4+2x^3+4x^2}-\sqrt{x^4+2x^3-x^2})}\\ & =\underset{x\to \mathrm{\infty }}{lim}(\sqrt{x^4+2x^3+4x^2}-\sqrt{x^4+2x^3-x^2})\\ & \times {\displaystyle \frac{\sqrt{x^4+2x^3+4x^2}+\sqrt{x^4+2x^3-x^2}}{\sqrt{x^4+2x^3+4x^2}+\sqrt{x^4+2x^3-x^2}}}\\ & =\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{(x^4+2x^3+4x^2)-(x^4+2x^3-x^2)}{\sqrt{x^4+2x^3+4x^2}+\sqrt{x^4+2x^3-x^2}}}\\ & =\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{5x^2}{\sqrt{x^4+2x^3+4x^2}+\sqrt{x^4+2x^3-x^2}}}\\ & \text{Bagi setiap suku dengan}{x}^2\\ & =\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{{\displaystyle \frac{5x^2}{x^2}}}{\sqrt{{\displaystyle \frac{x^4}{x^4}}+{\displaystyle \frac{2x^3}{x^4}}+{\displaystyle \frac{4x^2}{x^4}}}+\sqrt{{\displaystyle \frac{x^4}{x^4}}+{\displaystyle \frac{2x^3}{x^4}}-{\displaystyle \frac{x^2}{x^4}}}}}\\ & ={\displaystyle \frac{5}{\sqrt{1+0+0}+\sqrt{1+0-0}}}={\displaystyle \frac{5}{2}}.\end{array}$$Jadi, nilai dari limit tersebut adalah $\boxed{{\displaystyle {\displaystyle \frac{5}{2}}}}$
(Jawaban E)

Gunakan rumus
$$\boxed{{\displaystyle {\displaystyle \underset{x\to \mathrm{\infty }}{lim}(\sqrt{a{x}^2+bx+c}-\sqrt{a{x}^2+px+q})={\displaystyle \frac{b-p}{2\sqrt{a}}}}}}$$Untuk kasus ini, diketahui bahwa $a=9,b=5,$ dan p =-7.$
Dengan demikian, diperoleh
$$\begin{array}{rl}& {\displaystyle \underset{x\to \mathrm{\infty }}{lim}(\sqrt{9x^2+5x+5}-\sqrt{9x^2-7x-4})}\\ & ={\displaystyle \frac{5-(-7)}{2\sqrt{9}}}={\displaystyle \frac{12}{6}}=2.\end{array}$$Jadi, nilai dari limit tersebut adalah $\boxed{{\displaystyle 2}}$
(Jawaban D)

Perhatikan bahwa 
$\begin{array}{rl}3x+1& =\sqrt{(3x+1{)}^2}\\ & =\sqrt{9x^2+6x+1}\end{array}$
berlaku karena $x$ menuju takhingga (nilainya dipastikan positif). 
Untuk itu, dengan menggunakan rumus 
$$\boxed{{\displaystyle {\displaystyle \underset{x\to \mathrm{\infty }}{lim}(\sqrt{a{x}^2+bx+c}-\sqrt{a{x}^2+px+q})={\displaystyle \frac{b-p}{2\sqrt{a}}}}}}$$(Diketahui: $a=9,b=6,p=4$)
diperoleh
$${\displaystyle \underset{x\to \mathrm{\infty }}{lim}(3x+1)-\sqrt{9x^2+4x-7})={\displaystyle \frac{6-4}{2\sqrt{9}}}={\displaystyle \frac{1}{3}}.}$$Jadi, nilai dari limit tersebut adalah $\boxed{{\displaystyle {\displaystyle \frac{1}{3}}}}$
(Jawaban D)

Perhatikan bahwa bentuk $-x+2$ dapat ditulis menjadi
$\begin{array}{rl}-(x-2)& =-\sqrt{(x-2{)}^2}\\ & =-\sqrt{x^2-4x+4}.\end{array}$
Dengan demikian, diperoleh
${\displaystyle \underset{x\to \mathrm{\infty }}{lim}(\sqrt{x^2+3x+2}-(\sqrt{x^2-4x+4}).}$
Gunakan rumus 
$$\boxed{{\displaystyle {\displaystyle \underset{x\to \mathrm{\infty }}{lim}(\sqrt{a{x}^2+bx+c}-\sqrt{a{x}^2+px+q})={\displaystyle \frac{b-p}{2\sqrt{a}}}}}}$$untuk $a=1,b=3,p=-4$ sehingga diperoleh
$$\begin{array}{rl}& {\displaystyle \underset{x\to \mathrm{\infty }}{lim}(\sqrt{x^2+3x+2}-(\sqrt{x^2-4x+4})}\\ & ={\displaystyle \frac{3-(-4)}{2\sqrt{1}}}={\displaystyle \frac{7}{2}}=3,5.\end{array}$$Jadi, nilai dari limit tersebut adalah $\boxed{{\displaystyle 3,5}}$
(Jawaban B)

Dengan menggunakan sifat khusus limit takhingga dengan bentuk:
$${\displaystyle \underset{x\to \mathrm{\infty }}{lim}(\sqrt{a{x}^2+bx+c}-\sqrt{p{x}^2+qx+r})=\{\begin{array}{l}\mathrm{\infty },\text{jika}a>p\\ {\displaystyle \frac{b-q}{2\sqrt{a}}},\text{jika}a=p\\ -\mathrm{\infty },\text{jika}a<p\end{array}}$$diperoleh
$$\begin{array}{rl}& {\displaystyle \underset{x\to \mathrm{\infty }}{lim}(\sqrt{(2x-1)(x+2)}-(x\sqrt{2}+1))}\\ & =\underset{x\to \mathrm{\infty }}{lim}((\sqrt{2x^2+3x-2}-\sqrt{2x^2})-1)\\ & ={\displaystyle \frac{3-0}{2\sqrt{2}}}-1={\displaystyle \frac{3}{4}}\sqrt{2}-1.\end{array}$$Jadi, nilai dari $$\boxed{{\displaystyle {\displaystyle \underset{x\to \mathrm{\infty }}{lim}(\sqrt{(2x-1)(x+2)}-(x\sqrt{2}+1))={\displaystyle \frac{3}{4}}\sqrt{2}-1}}}$$(Jawaban B)

Ubah bentuk fungsinya sehingga membentuk selisih bentuk kuadrat dalam tanda akar agar nilai limitnya dapat langsung ditentukan.
$$\begin{array}{rl}& {\displaystyle \underset{x\to \mathrm{\infty }}{lim}[\sqrt{(x+a)(x+b)}-x]}\\ & =\underset{x\to \mathrm{\infty }}{lim}[\sqrt{x^2+(a+b)x+ab}-\sqrt{x^2}]\\ & ={\displaystyle \frac{(a+b)-0}{2\sqrt{1}}}={\displaystyle \frac{a+b}{2}}\end{array}$$Jadi, nilai dari $\boxed{{\displaystyle {\displaystyle \underset{x\to \mathrm{\infty }}{lim}[\sqrt{(x+a)(x+b)}-x]={\displaystyle \frac{a+b}{2}}}}}$
(Jawaban A)

Dengan menggunakan salah satu sifat akar:
$\boxed{{\displaystyle \sqrt{(a+b)\pm 2\sqrt{ab}}=\sqrt{a}\pm \sqrt{b}}}$
dan sejumlah sifat limit dasar, diperoleh
$$\begin{array}{rl}& {\displaystyle \underset{x\to \mathrm{\infty }}{lim}\sqrt{5(x-1)+2\sqrt{4x^2-23x-6}}}\\ & =\underset{x\to \mathrm{\infty }}{lim}\sqrt{[(4x+1)+(x-6)]+2\sqrt{(4x+1)(x-6)}}\\ & =\underset{x\to \mathrm{\infty }}{lim}(\sqrt{4x+1}+\sqrt{x-6})\\ & =\mathrm{\infty }.\end{array}$$Penjelasan pada langkah terakhir: Karena $x$ nilainya menuju takhingga, $\sqrt{4x+1}$ akan membesar nilainya, begitu juga dengan $\sqrt{x-6}$ sehingga bila dijumlahkan keduanya, hasilnya akan takhingga. 
(Jawaban E)

Dengan menggunakan metode pengalian akar sekawan, diperoleh
$$\begin{array}{rl}& {\displaystyle \underset{x\to \mathrm{\infty }}{lim}2x(\sqrt{9+{\displaystyle \frac{10}{x}}}-3)}\\ & =\underset{x\to \mathrm{\infty }}{lim}2x(\sqrt{9+{\displaystyle \frac{10}{x}}}-3)\times {\displaystyle \frac{\sqrt{9+{\displaystyle \frac{10}{x}}}+3}{\sqrt{9+{\displaystyle \frac{10}{x}}}+3}}\\ & =\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{2x\cdot (9+{\displaystyle \frac{10}{x}}-3^2)}{\sqrt{9+{\displaystyle \frac{10}{x}}}+3}}\\ & =\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{2\cdot 10}{\sqrt{9+{\displaystyle \frac{10}{x}}}+3}}\\ & ={\displaystyle \frac{20}{\sqrt{9+0}+3}}={\displaystyle \frac{10}{3}}.\end{array}$$Jadi, nilai dari $\boxed{{\displaystyle {\displaystyle \underset{x\to \mathrm{\infty }}{lim}2x(\sqrt{9+{\displaystyle \frac{10}{x}}}-3)={\displaystyle \frac{10}{3}}}}}$ 
(Jawaban A)

Perhatikan bahwa bentuk
${\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{(x-2)\sqrt{(x+2)+\sqrt{4x}}}{x\sqrt{2x}-2\sqrt{x}+2\sqrt{2}}}}$
dapat dinyatakan sebagai
$${\displaystyle \underset{x\to \mathrm{\infty }}{lim}\sqrt{{\left({\displaystyle \frac{(x-2)\sqrt{(x+2)+\sqrt{4x}}}{x\sqrt{2x}-2\sqrt{x}+2\sqrt{2}}}\right)}^2}.}$$Dengan demikian, diperoleh
$${\displaystyle \underset{x\to \mathrm{\infty }}{lim}\sqrt{{\displaystyle \frac{(x^2-4x+4)(x+2+\sqrt{4x})}{(\sqrt{2}{x}^{\frac{3}{2}}-2x^{\frac{1}{2}}+2\sqrt{2}{)}^2}}}.}$$Tinjau hanya pada variabel berpangkat tertingginya.
${\displaystyle \underset{x\to \mathrm{\infty }}{lim}\sqrt{{\displaystyle \frac{x^3+\cdots }{2x^3+\cdots }}}}$
Bagi setiap suku dengan $x^3.$
${\displaystyle \underset{x\to \mathrm{\infty }}{lim}\sqrt{{\displaystyle \frac{{\displaystyle \frac{x^3}{x^3}}+\cdots }{{\displaystyle \frac{2x^3}{x^3}}+\cdots }}}}$
Gunakan sifat limit takhingga untuk memperoleh
$\sqrt{{\displaystyle \frac{1+0}{2+0}}}={\displaystyle \frac{1}{\sqrt{2}}}={\displaystyle \frac{1}{2}}\sqrt{2}.$
Jadi, nilai dari $$\boxed{{\displaystyle {\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{(x-2)\sqrt{(x+2)+\sqrt{4x}}}{x\sqrt{2x}-2\sqrt{x}+2\sqrt{2}}}={\displaystyle \frac{1}{2}}\sqrt{2}}}}$$(Jawaban D)

$\begin{array}{rl}& {\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{3^{x+1}+2^x-3}{3^{x+2}-2^{x-1}+4}}}\\ & =\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{3^{x+1}+2^x-3}{3^{x+2}-2^{x-1}+4}}\times {\displaystyle \frac{\frac{1}{3^{x+2}}}{\frac{1}{3^{x+2}}}}\\ & =\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{\frac{1}{3}+\frac{2^x}{3^{x+2}}-{\displaystyle \frac{3}{3^{x+2}}}}{1-\frac{2^{x-1}}{3^{x+2}}+{\displaystyle \frac{4}{3^{x+2}}}}}\\ & ={\displaystyle \frac{\frac{1}{3}+0-0}{1-0+0}}={\displaystyle \frac{1}{3}}\end{array}$
Jadi, nilai dari $\boxed{{\displaystyle {\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{3^{x+1}+2^x-3}{3^{x+2}-2^{x-1}+4}}={\displaystyle \frac{1}{3}}}}}$
(Jawaban C)

Faktorkan $2^x$ keluar dari akar.
$$\begin{array}{rl}& {\displaystyle \underset{x\to \mathrm{\infty }}{lim}(\sqrt[3]{8^x+3^x}-\sqrt{4^x-2^x})}\\ & =\underset{x\to \mathrm{\infty }}{lim}(\sqrt[3]{8^x(1+{\left({\displaystyle \frac{3}{8}}\right)}^x)}-\sqrt{4^x(1-{\left({\displaystyle \frac{1}{2}}\right)}^x)})\\ & =\underset{x\to \mathrm{\infty }}{lim}(2^x\sqrt[3]{1+{\left({\displaystyle \frac{3}{8}}\right)}^x}-2^x\sqrt{1-{\left({\displaystyle \frac{1}{2}}\right)}^x})\end{array}$$Selanjutnya, dengan menggunakan Aproksimasi (Pendekatan) Binomial, diperoleh
$$\begin{array}{rl}& \underset{x\to \mathrm{\infty }}{lim}(2^x(1+{\displaystyle \frac{1}{3}}{\left({\displaystyle \frac{3}{8}}\right)}^x)-2^x(1-{\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{1}{2}}\right)}^x))\\ & =\underset{x\to \mathrm{\infty }}{lim}(\cancel{2^x}+{\displaystyle \frac{1}{3}}{\left({\displaystyle \frac{6}{8}}\right)}^x-\cancel{2^x}+{\displaystyle \frac{1}{2}})\\ & ={\displaystyle \frac{1}{3}}(0)+{\displaystyle \frac{1}{2}}={\displaystyle \frac{1}{2}}.\end{array}$$Jadi, nilai dari $\boxed{{\displaystyle {\displaystyle \underset{x\to \mathrm{\infty }}{lim}(\sqrt[3]{8^x+3^x}-\sqrt{4^x-2^x})={\displaystyle \frac{1}{2}}}}}$
(Jawaban D)

Alternatif 1: Membagi dengan Variabel Pangkat Tertinggi
Sebelumnya, perlu diketahui bahwa bentuk akar kuadrat dapat dirasionalkan dengan cara dikalikan akar sekawan, sedangkan bentuk akar kubik, seperti $\sqrt[3]{x}+a$ dirasionalkan dengan cara dikalikan $\sqrt[3]{x^2}+a\sqrt[3]{x}+a^2$ berdasarkan pemfaktoran $a^3-b^3=(a-b)(a^2+ab+b^2)$.
Oleh karena itu, kita peroleh
$$\begin{array}{rl}& {\displaystyle \underset{x\to \mathrm{\infty }}{lim}(\sqrt[3]{8x^3+12x^2-5}-\sqrt{x^2+6x}-x+3)}\\ & =\underset{x\to \mathrm{\infty }}{lim}(\sqrt[3]{8x^3+12x^2-5}-2x)-\underset{x\to \mathrm{\infty }}{lim}(\sqrt{x^2+6x}-x)+3\\ & =\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{(8x^3+12x^2-5)-(8x^3)}{(\sqrt[3]{8x^3+12x^2-5}{)}^2+2x\sqrt[3]{8x^3+12x^2-5}+4x^2}}-\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{(x^2+6x)-x^2}{\sqrt{x^2+6x}+x}}+3\\ & =\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{\cancel{x^2}(12-{\displaystyle \frac{5}{x^2}})}{\cancel{x^2}({((\sqrt[3]{8+{\displaystyle \frac{12}{x}}-{\displaystyle \frac{5}{x^3}}})}^2+2\sqrt[3]{8+{\displaystyle \frac{12}{x}}-{\displaystyle \frac{5}{x^3}}}+4)}}-\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{6}{\sqrt{1+{\displaystyle \frac{6}{x}}}+1}}+3\\ & ={\displaystyle \frac{12-0}{(\sqrt[3]{8+0-0}{)}^2+2\sqrt[3]{8+0-0}+4}}-{\displaystyle \frac{6}{\sqrt{1+0}+1}}+3\\ & ={\displaystyle \frac{12}{4+4+4}}-3+3=1.\end{array}$$Alternatif 2: Menggunakan Aproksimasi Binomial
Perhatikan bahwa ekspresi limit yang diberikan dapat kita tulis menjadi
$$\begin{array}{rl}& {\displaystyle \underset{x\to \mathrm{\infty }}{lim}(\sqrt[3]{8x^3+12x^2-5}-2x)-\underset{x\to \mathrm{\infty }}{lim}(\sqrt{x^2+6x}-x)+3}\\ & =\underset{x\to \mathrm{\infty }}{lim}2x(\sqrt[3]{1+{\displaystyle \frac{3}{2x}}-{\displaystyle \frac{5}{8x^3}}}-1)-\underset{x\to \mathrm{\infty }}{lim}x(\sqrt{1+{\displaystyle \frac{6}{x}}}-1)+3.\end{array}$$Dengan menggunakan Aproksimasi Binomial untuk akar kuadrat dan akar kubiknya, kita peroleh
$$\begin{array}{rl}& {\displaystyle \underset{x\to \mathrm{\infty }}{lim}2x((1+{\displaystyle \frac{1}{3}}({\displaystyle \frac{3}{2x}}-{\displaystyle \frac{5}{8x^3}}))-1)-\underset{x\to \mathrm{\infty }}{lim}x((1+{\displaystyle \frac{1}{2}}\cdot {\displaystyle \frac{6}{x}})-1)+3}\\ & =\underset{x\to \mathrm{\infty }}{lim}2x({\displaystyle \frac{1}{2x}}-{\displaystyle \frac{5}{24x^3}})-\underset{x\to \mathrm{\infty }}{lim}x\left({\displaystyle \frac{3}{x}}\right)+3\\ & =\underset{x\to \mathrm{\infty }}{lim}(1-{\displaystyle \frac{5}{12x^2}})-\underset{x\to \mathrm{\infty }}{lim}3+3\\ & =(1-0)-3+3=1.\end{array}$$Jadi, nilai dari limit tersebut sama dengan $\boxed{{\displaystyle 1}}$
(Jawaban C)

Sederhanakan rumus fungsinya terlebih dahulu dengan memanfaatkan rumus pemfaktoran $a^2-b^2=(a+b)(a-b)$ dan konsep teleskopik.
$$\begin{array}{rl}& (1-{\displaystyle \frac{1}{2^2}})(1-{\displaystyle \frac{1}{3^2}})(1-{\displaystyle \frac{1}{4^2}})\cdots (1-{\displaystyle \frac{1}{n^2}})\\ & =(1-{\displaystyle \frac{1}{2}})(1+{\displaystyle \frac{1}{2}})(1-{\displaystyle \frac{1}{3}})(1+{\displaystyle \frac{1}{3}})\\ & (1-{\displaystyle \frac{1}{4}})(1+{\displaystyle \frac{1}{4}})\cdots (1-{\displaystyle \frac{1}{n}})(1+{\displaystyle \frac{1}{n}})\\ & ={\displaystyle \frac{1}{2}}\cdot \cancel{{\displaystyle \frac{3}{2}}\cdot {\displaystyle \frac{2}{3}}\cdot {\displaystyle \frac{4}{3}}\cdot {\displaystyle \frac{3}{4}}\cdot {\displaystyle \frac{5}{4}}\cdots {\displaystyle \frac{n-1}{n}}}\cdot {\displaystyle \frac{n+1}{n}}\\ & ={\displaystyle \frac{1}{2}}\cdot {\displaystyle \frac{n+1}{n}}={\displaystyle \frac{n+1}{2n}}\end{array}$$Dengan demikian, dapat ditulis
$\begin{array}{rl}{\displaystyle \underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{n+1}{2n}}}& =\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{{\displaystyle \frac{n}{n}}+{\displaystyle \frac{1}{n}}}{{\displaystyle \frac{2n}{n}}}}\\ & ={\displaystyle \frac{1+0}{2}}={\displaystyle \frac{1}{2}}.\end{array}$
Jadi, hasil dari
$\boxed{{\displaystyle \begin{array}{rl}& {\displaystyle \underset{n\to \mathrm{\infty }}{lim}(1-{\displaystyle \frac{1}{2^2}})(1-{\displaystyle \frac{1}{3^2}})}\\ & (1-{\displaystyle \frac{1}{4^2}})\cdots (1-{\displaystyle \frac{1}{n^2}})={\displaystyle \frac{1}{2}}\end{array}}}$
(Jawaban D)

Gunakan sifat limit takhingga khusus.
$$\boxed{{\displaystyle \begin{array}{rl}& {\displaystyle \underset{x\to \mathrm{\infty }}{lim}(\sqrt{a{x}^2+bx+c}-\sqrt{a{x}^2+px+q})={\displaystyle \frac{b-p}{2\sqrt{a}}}}\\ & \underset{x\to \mathrm{\infty }}{lim}(\sqrt[3]{a{x}^3+b{x}^2+cx+d}-\sqrt[3]{a{x}^3+p{x}^2+qx+r})={\displaystyle \frac{b-p}{3\sqrt[3]{a^2}}}\end{array}}}$$Dengan demikian, kita peroleh
$$\begin{array}{rl}& {\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{\sqrt{9x^2+12x-1}-\sqrt{9x^2-24x+10}}{\sqrt[3]{x^3+8x^2+x}-\sqrt[3]{x^3-x^2+2x}}}}\\ & ={\displaystyle \frac{{\displaystyle \frac{12-(-24)}{2\sqrt{9}}}}{{\displaystyle \frac{8-(-1)}{3\sqrt[3]{1^2}}}}}={\displaystyle \frac{{\displaystyle \frac{36}{6}}}{{\displaystyle \frac{9}{3}}}}={\displaystyle \frac{6}{3}}=2.\end{array}$$Jadi, nilai dari $$\boxed{{\displaystyle {\displaystyle \underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{\sqrt{9x^2+12x-1}-\sqrt{9x^2-24x+10}}{\sqrt[3]{x^3+8x^2+x}-\sqrt[3]{x^3-x^2+2x}}}=2}}}$$(Jawaban D)

Misalkan $y={\displaystyle \frac{1}{x}}$, ekuivalen dengan $x={\displaystyle \frac{1}{y}}$. Jika $x\to \mathrm{\infty }$, maka $y\to 0$ sehingga ditulis
$$\begin{array}{rl}{\displaystyle \underset{x\to \mathrm{\infty }}{lim}(3x+\sin {\displaystyle \frac{1}{x}})}& =\underset{y\to 0}{lim}({\displaystyle \frac{3}{y}}+\sin y)\\ & =\underset{y\to 0}{lim}{\displaystyle \frac{3}{y}}+\underset{y\to 0}{lim}\sin y\\ & =\mathrm{\infty }+0=\mathrm{\infty }.\end{array}$$Jadi, nilai dari $\boxed{{\displaystyle {\displaystyle \underset{x\to \mathrm{\infty }}{lim}(3x+\sin {\displaystyle \frac{1}{x}})=\mathrm{\infty }}}}$
(Jawaban E)