Category: SOAL OLIMPIADE

Substitusi langsung $x=4$ mengakibatkan munculnya bentuk taktentu ${\displaystyle \frac{0}{0}}.$
Dengan menggunakan metode pengalian akar sekawan sekaligus metode pemfaktoran, diperoleh
$$\begin{array}{rl}& {\displaystyle \underset{x\to 4}{lim}{\displaystyle \frac{x^3-64}{\sqrt{x}-\sqrt{3\sqrt{x}-2}}}}\\ =& \underset{x\to 4}{lim}{\displaystyle \frac{x^3-64}{\sqrt{x}-\sqrt{3\sqrt{x}-2}}}\times {\displaystyle \frac{\sqrt{x}+\sqrt{3\sqrt{x}-2}}{\sqrt{x}+\sqrt{3\sqrt{x}-2}}}\\ =& \underset{x\to 4}{lim}{\displaystyle \frac{(x^3-64)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}{x-(3\sqrt{x}-2)}}\\ =& \underset{x\to 4}{lim}{\displaystyle \frac{(x-4)(x^2+4x+16)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}{(\sqrt{x}-2)(\sqrt{x}-1)}}\\ =& \underset{x\to 4}{lim}{\displaystyle \frac{\cancel{(\sqrt{x}-2)}(\sqrt{x}+2)(x^2+4x+16)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}{\cancel{(\sqrt{x}-2)}(\sqrt{x}-1)}}\\ =& \underset{x\to 4}{lim}{\displaystyle \frac{(\sqrt{x}+2)(x^2+4x+16)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}{\sqrt{x}-1}}\\ =& {\displaystyle \frac{(\sqrt{4}+2)(4^2+4(4)+16)(\sqrt{4}+\sqrt{3\sqrt{4}-2})}{\sqrt{4}-1}}\\ =& {\displaystyle \frac{(2+2)(16+16+16)(2+2)}{2-1}}\\ =& 4\times 48\times 4=768\end{array}$$Jadi, nilai dari $\boxed{{\displaystyle {\displaystyle \underset{x\to 4}{lim}{\displaystyle \frac{x^3-64}{\sqrt{x}-\sqrt{3\sqrt{x}-2}}}=768}}}.$
(Jawaban E)

Dengan menggunakan dalil L’Hospital, kita dapat menentukan persamaan yang melibatkan $a$ dan $b$ pada bentuk limit tersebut, dengan syarat substitusi langsung $x=4$ menghasilkan bentuk taktentu ${\displaystyle \frac{0}{0}}.$ Ini berarti,
${\displaystyle \frac{4^2+a\sqrt{4}+b}{4-4}}={\displaystyle \frac{2a+b+16}{0}}={\displaystyle \frac{0}{0}}.$
Jadi, diperoleh persamaan $2a+b=-16$.
Selanjutnya, terapkan dalil L’Hospital (turunkan terhadap variabel $x$)
$\begin{array}{rl}{\displaystyle \underset{x\to 4}{lim}{\displaystyle \frac{x^2+a\sqrt{x}+b}{x-4}}}& =5\\ \stackrel{\text{L’H}}{\Rightarrow }\underset{x\to 4}{lim}{\displaystyle \frac{2x+\frac{1}{2\sqrt{x}}a}{1}}& =5\\ 2(4)+{\displaystyle \frac{1}{2\sqrt{4}}}a& =5\\ 8+{\displaystyle \frac{1}{4}}a& =5\\ a& =(5-8)\times 4\\ a& =-12\end{array}$
Substitusi $a=-12$ pada persamaan $2a+b=-16.$
$2(-12)+b=-16\iff b=8.$
Jadi, nilai $\boxed{{\displaystyle a+b=-12+8=-4}}.$

Bentuk limit di atas dapat ditentukan dengan menggunakan Dalil L’Hospital, namun syaratnya ketika $x=b$ disubstitusikan menghasilkan bentuk taktentu ${\displaystyle \frac{0}{0}}$. 
Pada penyebut, jelas substitusi menghasilkan $0.$
Pada pembilang, 
$\begin{array}{rl}4-\sqrt{a(x+b)}& =0\\ 4-\sqrt{a(b+b)}& =0\\ -\sqrt{2ab}& =-4\\ ab& =8& & (★)\end{array}$
Terapkan dalil L’Hospital dengan cara menurunkan masing-masing pembilang dan penyebut terhadap variabel $x$. 
$\begin{array}{rl}{\displaystyle \underset{x\to b}{lim}{\displaystyle \frac{4-\sqrt{a(x+b)}}{b-x}}}& =b\\ \stackrel{\text{L’H}}{\Rightarrow }\underset{x\to b}{lim}{\displaystyle \frac{-\frac{1}{2}a(a(x+b){)}^{-\frac{1}{2}}}{-1}}& =b\\ \text{Substitusi}x=b& \\ {\displaystyle \frac{a}{2\sqrt{a(b+b)}}}& =b\\ a& =2b\sqrt{2ab}\\ a^2& =4b^2\cdot 2ab\\ a^2& =8a{b}^3\\ a(a-8b^3)& =0\end{array}$
Diperoleh $a=0$ (tidak memenuhi karena diberikan bahwa $a<0$) atau $a=8b^3$. 
Substitusi $a=8b^3$ pada $★$, 
$\begin{array}{rl}ab& =8\\ (8b^3)b& =8\\ b^4& =1\end{array}$
Diperoleh $b=1$ (tidak memenuhi karena diberikan bahwa $b<0$) atau $b=-1$ (memenuhi). 
Untuk $b=-1$, diperoleh $a=8b^3=8(-1{)}^3=-8$. 
Jadi, $\boxed{{\displaystyle a-b=-8-(-1)=-7}}.$
(Jawaban B)

Perhatikan bentuk ${\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{f(x)}{x^2}}=1}$. Karena memiliki nilai limit berhingga, maka substitusi langsung $x=0$ harus menghasilkan bentuk taktentu ${\displaystyle \frac{0}{0}}$. Ini mengimplikasikan
${\displaystyle \frac{{\displaystyle \underset{x\to 0}{lim}f(x)}}{{\displaystyle \underset{x\to 0}{lim}{x}^2}}}=1$
sehingga mengharuskan $\boxed{{\displaystyle {\displaystyle \underset{x\to 0}{lim}f(x)=0}}}.$
(Jawaban B)

Dengan menggunakan sifat dasar limit, persamaan ${\displaystyle \underset{x\to a}{lim}[f(x)-3g(x)]=2}$ dapat kita tuliskan menjadi
${\displaystyle \underset{x\to a}{lim}f(x)-3\underset{x\to a}{lim}g(x)=2}$
dan persamaan ${\displaystyle \underset{x\to a}{lim}[3f(x)+g(x)]=1}$ dapat kita tuliskan menjadi
${\displaystyle 3\underset{x\to a}{lim}f(x)+\underset{x\to a}{lim}g(x)=1.}$
Sekarang, misalkan ${\displaystyle \underset{x\to a}{lim}f(x)=m}$ dan ${\displaystyle \underset{x\to a}{lim}g(x)=n}$, sehingga terbentuk SPLDV:
$\{\begin{array}{l}m-3n=2\\ 3m+n=1\end{array}$
Selesaikan dengan metode gabungan. 
$\begin{array}{rl}\begin{array}{rl}m-3n& =2\\ 3m+n& =1\end{array}\left|\begin{array}{c}\times 1\\ \times 3\end{array}\right|& \begin{array}{rl}m-3n& =2\\ 9m+3n& =3\end{array}\\ & +\\ & \begin{array}{rl}10m& =5\\ m& ={\displaystyle \frac{5}{10}}={\displaystyle \frac{1}{2}}\end{array}\end{array}$
Untuk $m={\displaystyle \frac{1}{2}}$, diperoleh:
$\begin{array}{rl}3m+n& =1\\ n& =1-3m\\ & =1-3\left({\displaystyle \frac{1}{2}}\right)\\ & =-{\displaystyle \frac{1}{2}}\end{array}$
Ini berarti, $m={\displaystyle \underset{x\to a}{lim}f(x)={\displaystyle \frac{1}{2}}}$ dan $n={\displaystyle \underset{x\to a}{lim}g(x)=-{\displaystyle \frac{1}{2}}.}$ 
Dengan demikian, 
$\begin{array}{rl}{\displaystyle \underset{x\to a}{lim}f(x)g(x)}& =\underset{x\to a}{lim}f(x)\cdot \underset{x\to a}{lim}g(x)\\ & ={\displaystyle \frac{1}{2}}\cdot (-{\displaystyle \frac{1}{2}})=-{\displaystyle \frac{1}{4}}\end{array}$
Jadi, hasil dari $\boxed{{\displaystyle {\displaystyle \underset{x\to a}{lim}f(x)g(x)=-{\displaystyle \frac{1}{4}}}}}.$
(Jawaban B)

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

Dengan menggunakan substitusi langsung, diperoleh
$\begin{array}{rl}{\displaystyle \underset{x\to 2}{lim}{\displaystyle \frac{\frac{1}{3}A{x}^3+\frac{1}{2}B{x}^2-3x}{x^3-2x^2-9x+16}}}& ={\displaystyle \frac{3}{10}}\\ {\displaystyle \frac{\frac{1}{3}A(2{)}^3+\frac{1}{2}B(2{)}^2-3(2)}{(2{)}^3-2(2{)}^2-9(2)+16}}& ={\displaystyle \frac{3}{10}}\\ {\displaystyle \frac{\frac{8}{3}A+2B-6}{8-8-18+16}}& ={\displaystyle \frac{3}{10}}\\ {\displaystyle \frac{\frac{8}{3}A+2B-6}{-2}}\times {\displaystyle \frac{-5}{-5}}& ={\displaystyle \frac{3}{10}}\\ {\displaystyle \frac{-\frac{40}{3}A-10B+30}{\cancel{10}}}& ={\displaystyle \frac{3}{\cancel{10}}}\\ -\frac{40}{3}A-10B+30& =3\\ \text{Kalikan kedua ruas dengan}& (-3)\\ 40A+30B-90& =-9\\ 40A+30B& =81\end{array}$
Jadi, nilai dari $\boxed{{\displaystyle 40A+30B=81}}.$
(Jawaban B)

Gunakan metode pengalian akar sekawan dua kali. 
Kalikan fungsinya dengan ${\displaystyle \frac{\sqrt{7x^2+2\sqrt{x}}+\sqrt{11x^2-2\sqrt{x}}}{\sqrt{7x^2+2\sqrt{x}}+\sqrt{11x^2-2\sqrt{x}}}}$
sehingga didapat
$$\begin{array}{rl}& {\displaystyle \underset{x\to 1}{lim}{\displaystyle \frac{(\sqrt{3x^2+\sqrt{x}}-\sqrt{5x^2-\sqrt{x}})(\sqrt{7x^2+2\sqrt{x}})}{(7x^2+2\sqrt{x})-(11x^2-2\sqrt{x})}}}\\ & ={\displaystyle \underset{x\to 1}{lim}{\displaystyle \frac{(\sqrt{3x^2+\sqrt{x}}-\sqrt{5x^2-\sqrt{x}})(\sqrt{7x^2+2\sqrt{x}})}{-4x^2+4\sqrt{x}}}}\end{array}$$Selanjutnya, kalikan fungsinya dengan ${\displaystyle \frac{\sqrt{3x^2+\sqrt{x}}+\sqrt{5x^2-\sqrt{x}}}{\sqrt{3x^2+\sqrt{x}}+\sqrt{5x^2-\sqrt{x}}}}$
sehingga didapat
$$\begin{array}{rl}& {\displaystyle \underset{x\to 1}{lim}{\displaystyle \frac{\cancel{(-2x^2+2\sqrt{x})}(\sqrt{7x^2+2\sqrt{x}})+\sqrt{11x^2-2\sqrt{x}})}{{\cancel{(-4x^2+4\sqrt{x})}}^2(\sqrt{(3x^2+\sqrt{x}}+\sqrt{5x^2-\sqrt{x}})}}}\\ & =\underset{x\to 1}{lim}{\displaystyle \frac{\sqrt{7x^2+2\sqrt{x}}+\sqrt{11x^2-2\sqrt{x}}}{2(\sqrt{3x^2+\sqrt{x}}+\sqrt{5x^2-\sqrt{x}})}}\\ & ={\displaystyle \frac{\sqrt{7(1{)}^2+2\sqrt{1}}+\sqrt{11(1{)}^2-2\sqrt{1}}}{2(\sqrt{3(1{)}^2+\sqrt{1}}+\sqrt{5(1{)}^2-\sqrt{1}}}}\\ & ={\displaystyle \frac{3+3}{2(2+2)}}={\displaystyle \frac{6}{8}}={\displaystyle \frac{3}{4}}\end{array}$$Jadi, nilai dari $$\boxed{{\displaystyle {\displaystyle \underset{x\to 1}{lim}{\displaystyle \frac{\sqrt{3x^2+\sqrt{x}}-\sqrt{5x^2-\sqrt{x}}}{\sqrt{7x^2+2\sqrt{x}}-\sqrt{11x^2-2\sqrt{x}}}}={\displaystyle \frac{3}{4}}}}}.$$(Jawaban E)

Tinjau ${\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{x}{\sqrt{a+x}-\sqrt{a-x}}}=b}$.
Dengan menggunakan metode pengalian akar sekawan, didapat
$$\begin{array}{rl}{\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{x}{\sqrt{a+x}-\sqrt{a-x}}}}& =b\\ \underset{x\to 0}{lim}{\displaystyle \frac{x}{\sqrt{a+x}-\sqrt{a-x}}}& \times {\displaystyle \frac{\sqrt{a+x}+\sqrt{a-x}}{\sqrt{a+x}+\sqrt{a-x}}}=b\\ \underset{x\to 0}{lim}{\displaystyle \frac{x(\sqrt{a+x}+\sqrt{a-x})}{(a+x)-(a-x)}}& =b\\ \underset{x\to 0}{lim}{\displaystyle \frac{\cancel{x}(\sqrt{a+x}+\sqrt{a-x})}{2\cancel{x}}}& =b\\ \underset{x\to 0}{lim}{\displaystyle \frac{\sqrt{a+x}+\sqrt{a-x}}{2}}& =b\\ {\displaystyle \frac{\sqrt{a+0}+\sqrt{a-0}}{2}}& =b\\ 2\sqrt{a}& =2b\\ a^{\frac{1}{2}}& =b\end{array}$$Selanjutnya, tinjau ${\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{\sqrt{b+x}-\sqrt{b-x}}{x}}.}$
Dengan menggunakan metode pengalian akar sekawan, diperoleh
$$\begin{array}{rl}& {\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{\sqrt{b+x}-\sqrt{b-x}}{x}}}\\ =& \underset{x\to 0}{lim}{\displaystyle \frac{\sqrt{b+x}-\sqrt{b-x}}{x}}\times {\displaystyle \frac{\sqrt{b+x}+\sqrt{b-x}}{\sqrt{b+x}+\sqrt{b-x}}}\\ =& \underset{x\to 0}{lim}{\displaystyle \frac{(b+x)-(b-x)}{x(\sqrt{b+x}+\sqrt{b-x})}}\\ =& \underset{x\to 0}{lim}{\displaystyle \frac{2\cancel{x}}{\cancel{x}(\sqrt{b+x}+\sqrt{b-x})}}\\ =& \underset{x\to 0}{lim}{\displaystyle \frac{2}{\sqrt{b+x}+\sqrt{b-x}}}\\ =& {\displaystyle \frac{2}{\sqrt{b+0}+\sqrt{b-0}}}\\ =& {\displaystyle \frac{2}{2\sqrt{b}}}={\displaystyle \frac{1}{\sqrt{b}}}\end{array}$$Substitusikan nilai $b=a^{\frac{1}{2}}$ sehingga diperoleh
${\displaystyle \frac{1}{\sqrt{b}}}=b^{-\frac{1}{2}}=(a^{\frac{1}{2}}{)}^{-\frac{1}{2}}=a^{-\frac{1}{4}}.$
Jadi, nilai dari ${\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{\sqrt{b+x}-\sqrt{b-x}}{x}}=a^{-\frac{1}{4}}.}$
(Jawaban B)

Diketahui: ${\displaystyle \underset{x\to 1}{lim}{\displaystyle \frac{\sqrt{a{x}^4+b}-2}{x-1}}=A.}$
Dengan menggunakan sejumlah sifat limit dasar, diperoleh
$$\begin{array}{rl}& {\displaystyle \underset{x\to 1}{lim}{\displaystyle \frac{\sqrt{a{x}^4+b}-2x}{x^2+2x-3}}}\\ & =\underset{x\to 1}{lim}{\displaystyle \frac{\sqrt{a{x}^4+b}-2-2x+2}{(x+3)(x-1)}}\\ & =\underset{x\to 1}{lim}{\displaystyle \frac{1}{x+3}}\cdot \underset{x\to 1}{lim}({\displaystyle \frac{\sqrt{a{x}^4+b}-2}{x-1}}-{\displaystyle \frac{2x-2}{x-1}})\\ & =\underset{x\to 1}{lim}{\displaystyle \frac{1}{x+3}}\cdot \underset{x\to 1}{lim}({\displaystyle \frac{\sqrt{a{x}^4+b}-2}{x-1}}-{\displaystyle \frac{2\cancel{(x-1)}}{\cancel{x-1}}})\\ & ={\displaystyle \frac{1}{1+3}}\cdot (A-2)\\ & ={\displaystyle \frac{1}{4}}(A-2)\end{array}$$Jadi, nilai dari $\boxed{{\displaystyle {\displaystyle \underset{x\to 1}{lim}{\displaystyle \frac{\sqrt{a{x}^4+b}-2x}{x^2+2x-3}}={\displaystyle \frac{1}{4}}(A-2)}}}.$
(Jawaban B)

Dari persamaan ${\displaystyle \underset{x\to 2}{lim}{\displaystyle \frac{\sqrt{a{x}^2+b}-8}{x-2}}=A,}$ kita ketahui bahwa limitnya ada, sehingga substitusi $x=2$ pada bentuk ${\displaystyle \frac{\sqrt{a{x}^2+b}-8}{x-2}}$ seharusnya menghasilkan bentuk taktentu ${\displaystyle \frac{0}{0}}$, ditulis
${\displaystyle \frac{\sqrt{a(2{)}^2+b}-8}{2-2}}={\displaystyle \frac{\sqrt{4a+b}-8}{0}}={\displaystyle \frac{0}{0}}$
Jadi, diperoleh persamaan
$\begin{array}{rl}\sqrt{4a+b}-8& =0\\ \sqrt{4a+b}& =8\\ \text{Kuadratkan}& \text{kedua ruas}\\ 4a+b& =64\end{array}$
Sekarang, dapat kita tulis
$$\begin{array}{rl}{\displaystyle \underset{x\to 2}{lim}{\displaystyle \frac{\sqrt[3]{a{x}^2+b}-2x}{x^2+x-2}}}& ={\displaystyle \frac{\sqrt[3]{a(2{)}^2+b}-2(2)}{(2{)}^2+2-2}}\\ & ={\displaystyle \frac{\sqrt[3]{4a+b}-4}{4}}\\ & ={\displaystyle \frac{\sqrt[3]{64}-4}{4}}\\ & ={\displaystyle \frac{4-4}{4}}=0\end{array}$$Jadi, hasil dari $\boxed{{\displaystyle {\displaystyle \underset{x\to 2}{lim}{\displaystyle \frac{\sqrt[3]{a{x}^2+b}-2x}{x^2+x-2}}=0}}}.$
(Jawaban C)

Gunakan sifat pemfaktoran berikut.
$\boxed{{\displaystyle a^2-b^2=(a+b)(a-b)}}.$
Kita mulai dari bentuk limit yang ditanya, lalu kita arahkan supaya muncul bentuk limit yang diketahui.
$$\begin{array}{rl}& {\displaystyle \underset{t\to a}{lim}{\displaystyle \frac{(|a|-1{)}^4-(|t|-1{)}^4}{t-a}}}\\ & =\underset{t\to a}{lim}{\displaystyle \frac{[(|a|-1{)}^2-(|t|-1{)}^2]\cdot [(|a|-1{)}^2+(|t|-1{)}^2]}{(t-a)(t+a)}}\cdot (t+a)\\ & =-\underset{t\to a}{lim}{\displaystyle \frac{(|t|-1{)}^2-(|a|-1{)}^2}{t^2-a^2}}\cdot \underset{t\to a}{lim}[(t+a)\cdot ((|a|-1{)}^2+(|t|-1{)}^2)]\\ & =-M\cdot (a+a)((|a|-1{)}^2+(|a|-1{)}^2)\\ & =-M\cdot (2a)\cdot 2(|a|-1{)}^2\\ & =-4aM(|a|-1{)}^2\end{array}$$Jadi, nilai dari $$\boxed{{\displaystyle {\displaystyle \underset{t\to a}{lim}{\displaystyle \frac{(|a|-1{)}^4-(|t|-1{)}^4}{t-a}}=-4aM(|a|-1{)}^2}}}.$$(Jawaban C)

Karena fungsi ${\displaystyle \frac{x^2-x-b}{2-x}}$ memiliki nilai limit untuk $x$ mendekati $2$, maka substitusi langsung $x=2$ harus menghasilkan bentuk taktentu ${\displaystyle \frac{0}{0}}$ sehingga ditulis
${\displaystyle \frac{(2{)}^2-2-b}{2-2}}={\displaystyle \frac{2-b}{2-2}}={\displaystyle \frac{0}{0}}.$
Dengan demikian, diperoleh $b=2$. 
Selanjutnya, dapat ditentukan nilai $a$ sebagai berikut. 
$$\begin{array}{rl}{\displaystyle \underset{x\to 2}{lim}{\displaystyle \frac{x^2-x-2}{2-x}}}& =\underset{x\to 2}{lim}{\displaystyle \frac{\cancel{(x-2)}(x+1)}{-\cancel{(x-2)}}}\\ & =\underset{x\to 2}{lim}{\displaystyle \frac{x+1}{-1}}\\ & =-(2+1)=-3\end{array}$$Diperoleh nilai $a=-3.$
Jadi, hasil dari $\boxed{{\displaystyle b-a=2-(-3)=5}}.$
(Jawaban E)

Agar fungsi tersebut memiliki nilai limit ketika $x$ mendekati $2$, substitusi $x=2$ harus membuat nilai fungsinya menjadi ${\displaystyle \frac{0}{0}}$ (bentuk taktentu).
Pada pembilang, jelas $(2{)}^2-4=0$.
Pada penyebut,
$\sqrt{2p+q}-2=0\Rightarrow 2p+q=4.$
Selanjutnya, dengan menggunakan dalil L’Hospital pada bentuk limitnya, diperoleh
$\begin{array}{rl}{\displaystyle \underset{x\to 2}{lim}{\displaystyle \frac{x^2-4}{\sqrt{px+q}-2}}}& =8\\ \stackrel{\text{L’H}}{\Rightarrow }\underset{x\to 2}{lim}{\displaystyle \frac{2x}{{\displaystyle \frac{p}{2\sqrt{px+q}}}}}& =8\\ {\displaystyle \frac{2(2)}{{\displaystyle \frac{p}{2\sqrt{2p+q}}}}}& =8\\ 4(2\sqrt{2p+q})& =8p\\ \text{Substitusi}& 2p+q=4\\ 4(2\sqrt{4})& =8p\\ p& =2\end{array}$
Untuk itu, $2p+q=4\Rightarrow 2(2)+q=4\iff q=0$.
Jadi, $\boxed{{\displaystyle 3p-5q=3(2)-5(0)=6}}.$
(Jawaban E)

Agar ${\displaystyle \underset{x\to 2}{lim}f(x)}$ memiliki nilai, maka limit kiri dan limit kanannya harus sama. 
$\begin{array}{rl}{\displaystyle \underset{x\to 2^{-}}{lim}f(x)}& =\underset{x\to 2^{+}}{lim}f(x)\\ \underset{x\to 2}{lim}(3x-p)& =\underset{x\to 2}{lim}(2x+1)\\ 3(2)-p& =2(2)+1\\ 6-p& =5\\ p& =1\end{array}$
Jadi, nilai $p$ adalah $\boxed{{\displaystyle 1}}.$
(Jawaban A)

Dengan metode pengalian akar sekawan, diperoleh
$$\begin{array}{rl}{\displaystyle }& \underset{x\to 0}{lim}{\displaystyle \frac{\sqrt{5x^5+4{\sin }^{4}x}}{\sqrt{x^2+1}-1}}\\ =& \underset{x\to 0}{lim}{\displaystyle \frac{\sqrt{x^4(5x+4{\displaystyle \frac{{\sin }^{4}x}{x^4}})}}{\sqrt{x^2+1}-1}}\times {\displaystyle \frac{\sqrt{x^2+1}+1}{\sqrt{x^2+1}+1}}\\ =& \underset{x\to 0}{lim}{\displaystyle \frac{x^2\sqrt{5x+4{\displaystyle \frac{{\sin }^{4}x}{x^4}}}\cdot (\sqrt{x^2+1}+1)}{(x^2+1)-1}}\\ =& \underset{x\to 0}{lim}{\displaystyle \frac{x^2\sqrt{5x+4{\displaystyle \frac{{\sin }^{4}x}{x^4}}}\cdot (\sqrt{x^2+1}+1)}{x^2}}\times {\displaystyle \frac{{\displaystyle \frac{1}{x^2}}}{{\displaystyle \frac{1}{x^2}}}}\\ =& \underset{x\to 0}{lim}{\displaystyle \frac{\sqrt{5x+4{\displaystyle \frac{{\sin }^{4}x}{x^4}}}\cdot (\sqrt{x^2+1}+1)}{1}}\\ =& \sqrt{5(0)+4}\cdot (\sqrt{0^2+1}+1)\\ =& \sqrt{4}(\sqrt{1}+1)=4\end{array}$$Jadi, nilai dari $\boxed{{\displaystyle {\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{\sqrt{5x^5+4{\sin }^{4}x}}{\sqrt{x^2+1}-1}}=4}}}.$
(Jawaban B)

Dengan menggunakan dalil L’Hospital, akan ditentukan nilai limitnya sebagai berikut. Ingat bahwa turunannya terhadap variabel $p$, sehingga $x$ dianggap sebagai konstanta.
$$\begin{array}{rl}& {\displaystyle \underset{p\to 0}{lim}{\displaystyle \frac{f(x+2p)-f(x)}{2p}}}\\ & =\underset{p\to 0}{lim}{\displaystyle \frac{{\sin }^{2}3(x+2p)-{\sin }^{2}3x}{2p}}\\ & =\underset{p\to 0}{lim}{\displaystyle \frac{{\sin }^2(3x+6p)-{\sin }^{2}3x}{2p}}\\ & =\underset{p\to 0}{lim}{\displaystyle \frac{6\cdot 2\sin (3x+6p)\cos (3x+6p)-0}{2}}\\ & =6\sin (3x+0)\cos (3x+0)\\ & =6\sin 3x\cos 3x\end{array}$$Jadi, nilai dari limitnya adalah $\boxed{{\displaystyle 6\sin 3x\cos 3x}}.$
(Jawaban D)

Misalkan $y=x-1$. 
Untuk $x\to 1$, maka $y\to 0.$
Dengan demikian, limit di atas dapat ditulis kembali menjadi
$\begin{array}{rl}& {\displaystyle \underset{x\to 1}{lim}\sqrt{{\displaystyle \frac{\tan \pi (x-1)-(x+1)(x-1)}{-4(x-1)+\sin 2\pi (x-1)}}}}\\ & =\sqrt{\underset{y\to 0}{lim}{\displaystyle \frac{\tan \pi y-(y+2)y}{-4y+\sin 2\pi y}}}\\ & =\sqrt{\underset{y\to 0}{lim}{\displaystyle \frac{\frac{\tan \pi y}{y}-\frac{(y+2)y}{y}}{\frac{-4y}{y}+\frac{\sin 2\pi y}{y}}}}\\ & =\sqrt{{\displaystyle \frac{\pi -(0+2)}{-4+2\pi }}}\\ & =\sqrt{{\displaystyle \frac{\cancel{\pi -2}}{2\cancel{(\pi -2)}}}}\\ & =\sqrt{{\displaystyle \frac{1}{2}}}={\displaystyle \frac{1}{2}}\sqrt{2}\end{array}$
Jadi, nilai dari $$\boxed{{\displaystyle {\displaystyle \underset{x\to 1}{lim}\sqrt{{\displaystyle \frac{\tan (\pi x-\pi )-(x^2-1)}{(4-4x)+\sin 2(\pi x-\pi )}}}={\displaystyle \frac{1}{2}}\sqrt{2}}}}.$$(Jawaban C)

Diketahui ${\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{\cos x-1}{ax\sin x+b}}=1}$.
Karena limitnya ada, maka substitusi $x=0$ pada fungsi seharusnya menghasilkan bentuk taktentu ${\displaystyle \frac{0}{0}}$.
Kita tuliskan:
$\begin{array}{rl}{\displaystyle \frac{\cos 0-1}{a(0)\sin 0+b}}& ={\displaystyle \frac{0}{0}}\\ {\displaystyle \frac{1-1}{0+b}}& ={\displaystyle \frac{0}{0}}\\ {\displaystyle \frac{0}{0+b}}& ={\displaystyle \frac{0}{0}}\end{array}$
Kita peroleh $b=0$ agar persamaan terpenuhi.
Sekarang, persamaan limitnya dapat ditulis menjadi ${\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{\cos x-1}{ax\sin x}}=1.}$
Dengan mengalikan pembilang dan penyebutnya dengan $(\cos x+1)$, kita peroleh
$$\begin{array}{rl}{\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{\cos x-1}{ax\sin x}}{\displaystyle \frac{\cos x+1}{\cos x+1}}}& =1\\ \underset{x\to 0}{lim}{\displaystyle \frac{{\cos }^{2}x-1}{ax\sin x(\cos x+1)}}& =1\\ -{\displaystyle \frac{1}{a}}\underset{x\to 0}{lim}{\displaystyle \frac{{\sin }^{2}x}{x\sin x(\cos x+1)}}& =1\\ -{\displaystyle \frac{1}{a}}\cdot \underset{x\to 0}{lim}{\displaystyle \frac{\sin x}{x}}\cdot \underset{x\to 0}{lim}{\displaystyle \frac{\sin x}{\sin x}}\cdot \underset{x\to 0}{lim}{\displaystyle \frac{1}{\cos x+1}}& =1\\ -{\displaystyle \frac{1}{a}}\cdot 1\cdot 1\cdot {\displaystyle \frac{1}{\cos 0+1}}& =1\\ -{\displaystyle \frac{1}{2a}}& =1\\ a& =-{\displaystyle \frac{1}{2}}\end{array}$$Catatan: ${\sin }^{2}x=1-{\cos }^{2}x$
Jadi, nilai $\boxed{{\displaystyle a+b=-{\displaystyle \frac{1}{2}}+0=-{\displaystyle \frac{1}{2}}}}.$
(Jawaban C)