Soal-Soal Matematika/Limit

Rumus umum

$\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}$

dengan bentuk tak tentu: $\frac{0}{0},0^0,0\cdot \mathrm{\infty },\frac{\mathrm{\infty }}{\mathrm{\infty }},\mathrm{\infty }-\mathrm{\infty },1^{\mathrm{\infty }},{\mathrm{\infty }}^0$

Sifat limit

$\begin{array}{cccc}\underset{x\to p}{lim}& k& =& k\\ \underset{x\to p}{lim}& x& =& p\\ \underset{x\to p}{lim}& f(x)& =& f(p)\\ \underset{x\to p}{lim}& k\cdot f(x)& =& k\cdot \underset{x\to p}{lim}& f(x)\\ \underset{x\to p}{lim}& (f(x)+g(x))& =& \underset{x\to p}{lim}f(x)+\underset{x\to p}{lim}g(x)\\ \underset{x\to p}{lim}& (f(x)-g(x))& =& \underset{x\to p}{lim}f(x)-\underset{x\to p}{lim}g(x)\\ \underset{x\to p}{lim}& (f(x)\cdot g(x))& =& \underset{x\to p}{lim}f(x)\cdot \underset{x\to p}{lim}g(x)\\ \underset{x\to p}{lim}& (f(x)/g(x))& =& \underset{x\to p}{lim}f(x)/\underset{x\to p}{lim}g(x)\\ \underset{x\to p}{lim}& (f(x){)}^n& =& (\underset{x\to p}{lim}f(x){)}^n\\ \underset{x\to p}{lim}& \sqrt[n]{f(x)}& =& \sqrt[n]{\underset{x\to p}{lim}f(x)}\end{array}$

Rumus lainnya

$\begin{array}{ccc}\underset{x\to 0}{lim}& \frac{x}{\sin x}& =1\\ \underset{x\to 0}{lim}& \frac{x}{\tan x}& =1\\ \underset{x\to 0}{lim}& \frac{\sin x}{x}& =1\\ \underset{x\to 0}{lim}& \frac{\tan x}{x}& =1\\ \underset{x\to \mathrm{\infty }}{lim}& x\sin (\frac{1}{x})& =1\\ \underset{x\to \mathrm{\infty }}{lim}& x\tan (\frac{1}{x})& =1\\ \underset{x\to 0}{lim}& \frac{ax}{\sin bx}& =\frac{a}{b}\\ \underset{x\to 0}{lim}& \frac{ax}{\tan bx}& =\frac{a}{b}\\ \underset{x\to 0}{lim}& \frac{\sin ax}{bx}& =\frac{a}{b}\\ \underset{x\to 0}{lim}& \frac{\tan ax}{bx}& =\frac{a}{b}\\ \underset{x\to \mathrm{\infty }}{lim}& p^x& =0,-1<p<1\\ \underset{x\to \mathrm{\infty }}{lim}& \frac{a{x}^m+b}{p{x}^n+q}& =\frac{a}{p},m=n\\ \underset{x\to 0}{lim}& \frac{\frac{a}{x^m}+b}{\frac{p}{x^n}+q}& =\frac{a}{p},m=n\\ \underset{x\to \mathrm{\infty }}{lim}& \sqrt{a{x}^2+bx+c}-\sqrt{p{x}^2+qx+r}& =\frac{b-q}{2\sqrt{a}},a=p\\ \underset{x\to \mathrm{\infty }}{lim}& \sqrt[3]{a{x}^3+b{x}^2+cx+d}-\sqrt[3]{p{x}^3+q{x}^2+rx+s}& =\frac{b-q}{3\sqrt[3]{a^2}},a=p\\ \underset{x\to \mathrm{\infty }}{lim}& (1+\frac{1}{x}{)}^x& =e\\ \underset{x\to 0}{lim}& (1+x{)}^{\frac{1}{x}}& =e\\ \underset{x\to \mathrm{\infty }}{lim}& (1+\frac{a}{x}{)}^{bx}& =e^{ab}\\ \underset{x\to 0}{lim}& (1+ax{)}^{\frac{b}{x}}& =e^{ab}\\ \underset{x\to 0}{lim}& \frac{ln(1+x)}{x}& =1\\ \underset{x\to 0}{lim}& \frac{log(1+x)}{x}& =1\\ \underset{x\to 0}{lim}& \frac{e^x-1}{x}& =1\\ \underset{x\to 0}{lim}& \frac{a^x-1}{x}& =lna,a>0\\ \underset{x\to 1}{lim}& \frac{x^n-1}{x-1}& =n\end{array}$

Aturan L'Hôpital

$\underset{x\to p}{lim}\frac{f(x)}{g(x)}=\underset{x\to p}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}$

Aturan limit mutlak

Jika $\underset{x\to p}{lim}\frac{|x-p|}{x-p}$ maka hasilnya tidak ada karena batasan limit harus ada + atau -.

Jika $\underset{x\to p^{+}}{lim}\frac{|x-p|}{x-p}$ atau $\underset{x\to p^{-}}{lim}\frac{|x-p|}{x-p}$ maka hasilnya ada bila limit berlaku.

contoh soal

1. tentukan nilai limit dari

• $\underset{x\to 3}{lim}\frac{x^2-5x+6}{x^2-9}$
• $\underset{x\to p}{lim}\frac{x^2-(3+p)x+3p}{x^2+(4-p)x-4p}$
• $\underset{x\to 9}{lim}\frac{\sqrt{x}-3}{x^2-10x+9}$
• $\underset{x\to 8}{lim}\frac{x^2-10x+16}{\sqrt[3]{x}-2}$
• $\underset{x\to 4}{lim}\frac{(\sqrt{x}-2)(5x+4)}{x^2-5x+4}$
• $\underset{x\to 1}{lim}\frac{(x-1{)}^2}{\sqrt[3]{x^2}-2\sqrt[3]{x}+1}$
• $\underset{x\to 3}{lim}\frac{\frac{1}{3}-\frac{1}{2x-3}}{x-3}$
• $\underset{x\to 0}{lim}\frac{1-cosx}{8x^2}$
• $\underset{x\to b}{lim}\frac{xsinb-bsinx}{x-b}$
• $\underset{x\to \mathrm{\infty }}{lim}\frac{16x^5+12x+3}{4x^5+108x^3+67x}$
• $\underset{x\to \mathrm{\infty }}{lim}\frac{2^{5x+4}-5^{2x}}{2^{5x+3}+7^x}$
• $\underset{x\to 0}{lim}\frac{4e^{3x}-5(2{)}^{9x}+1}{3x}$
• $\underset{x\to 0}{lim}\frac{9\cdot 15^{3x}-3^{3x+2}}{27x}$
• $\underset{x\to \mathrm{\infty }}{lim}\sqrt{9x^2+12x+7}+\sqrt{4x^2+8x+3}-5x$
• $\underset{x\to \mathrm{\infty }}{lim}(\frac{n+1}{n-3}{)}^{\frac{3n-9}{4}}$
• $\underset{x\to \mathrm{\infty }}{lim}(\frac{n^2+4n+10}{n^2-4n+7}{)}^{\frac{2n^2-8n+14}{24n+9}}$
• $\underset{x\to \mathrm{\infty }}{lim}(\frac{3n+8}{3n+7}{)}^{6n-5}$
• $\underset{x\to \mathrm{\infty }}{lim}(\frac{4n+3}{3n+1}{)}^{\frac{3n+1}{2n+5}}$
• $\underset{x\to \mathrm{\infty }}{lim}\frac{5n^2}{1+3+5+\dots +n}$
• $\underset{x\to 5^{+}}{lim}\frac{x^2-x+20}{|x-5|}$
• $\underset{x\to -6^{-}}{lim}\frac{|x+6|}{x^2+2x-24}$
• $\underset{x\to 0}{lim}\frac{40^{2x}-8^{2x}-5^{2x}+1}{4x^2}$
• $\underset{x\to 4}{lim}\frac{x^2-5x+4}{x^2-x-12}$ (aturan L'Hôpital)

Jawab

• $\underset{x\to 3}{lim}\frac{x^2-5x+6}{x^2-9}=\underset{x\to 3}{lim}\frac{(x-3)(x-2)}{(x-3)(x+3)}=\underset{x\to 3}{lim}\frac{x-2}{x+3}=\frac{3-2}{3+3}=\frac{1}{6}$
• $\underset{x\to p}{lim}\frac{x^2-(3+p)x+3p}{x^2+(4-p)x-4p}=\underset{x\to p}{lim}\frac{(x-3)(x-p)}{(x+4)(x-p)}=\underset{x\to p}{lim}\frac{x-3}{x+4}=\frac{p-3}{p+4}$
• $\underset{x\to 9}{lim}\frac{\sqrt{x}-3}{x^2-10x+9}=\underset{x\to 9}{lim}\frac{(\sqrt{x}-3)(\sqrt{x}+3)}{(x-1)(x-9)(\sqrt{x}+3)}=\underset{x\to 9}{lim}\frac{x-9}{(x-1)(x-9)(\sqrt{x}+3)}=\underset{x\to 9}{lim}\frac{1}{(x-1)(\sqrt{x}+3)}=\frac{1}{(9-1)(\sqrt{9}+3)}=\frac{1}{48}$
• $\underset{x\to 8}{lim}\frac{x^2-10x+16}{\sqrt[3]{x}-2}=\underset{x\to 8}{lim}\frac{(x-8)(x-2)(\sqrt[3]{x^2}+2\sqrt[3]{x}+4)}{(\sqrt[3]{x}-2)(\sqrt[3]{x^2}+2\sqrt[3]{x}+4)}=\underset{x\to 8}{lim}\frac{(x-8)(x-2)(\sqrt[3]{x^2}+2\sqrt[3]{x}+4)}{x-8}=\underset{x\to 8}{lim}(x-2)(\sqrt[3]{x^2}+2\sqrt[3]{x}+4)=(8-2)(\sqrt[3]{8^2}+2\sqrt[3]{8}+4)=6(4+4+4)=72$
• $\underset{x\to 4}{lim}\frac{(\sqrt{x}-2)(5x+4)}{x^2-5x+4}=\underset{x\to 4}{lim}\frac{(\sqrt{x}-2)(5x+4)}{(x-4)(x-1)}=\underset{x\to 4}{lim}\frac{(\sqrt{x}-2)(5x+4)}{(\sqrt{x}-2)(\sqrt{x}+2)(x-1)}=\underset{x\to 4}{lim}\frac{5x+4}{(\sqrt{x}+2)(x-1)}=\frac{5(4)+4}{(\sqrt{4}+2)(4-1)}=2$
• $\underset{x\to 1}{lim}\frac{(x-1{)}^2}{\sqrt[3]{x^2}-2\sqrt[3]{x}+1}=\underset{x\to 1}{lim}\frac{({\sqrt[3]{x}}^3-1^3{)}^2}{\sqrt[3]{x^2}-2\sqrt[3]{x}+1}=\underset{x\to 1}{lim}\frac{({\sqrt[3]{x}}^3-1^3{)}^2}{(\sqrt[3]{x}-1{)}^2}=\underset{x\to 1}{lim}\frac{((\sqrt[3]{x}-1)(\sqrt[3]{x^2}+\sqrt[3]{x}+1){)}^2}{(\sqrt[3]{x}-1{)}^2}=\underset{x\to 1}{lim}\frac{(\sqrt[3]{x}-1{)}^2(\sqrt[3]{x^2}+\sqrt[3]{x}+1{)}^2}{(\sqrt[3]{x}-1{)}^2}=\underset{x\to 1}{lim}(\sqrt[3]{x^2}+\sqrt[3]{x}+1{)}^2=(\sqrt[3]{1^2}+\sqrt[3]{1}+1{)}^2=(1+1+1{)}^2=9$
• $\underset{x\to 3}{lim}\frac{\frac{1}{3}-\frac{1}{2x-3}}{x-3}=\underset{x\to 3}{lim}\frac{\frac{2x-3-3}{3(2x-3)}}{x-3}=\underset{x\to 3}{lim}\frac{\frac{2x-6}{3(2x-3)}}{x-3}=\underset{x\to 3}{lim}\frac{2(x-3)}{3(2x-3)(x-3)}=\underset{x\to 3}{lim}\frac{2}{3(2x-3)}=\frac{2}{3(2(3)-3)}=\frac{2}{3(3)}=\frac{2}{9}$
• $\underset{x\to 0}{lim}\frac{1-cosx}{8x^2}=\underset{x\to 0}{lim}\frac{2si{n}^2\frac{x}{2}}{8x^2}=\frac{2sin\frac{x}{2}sin\frac{x}{2}}{8xx}=\frac{2}{8}\cdot \frac{1}{2}\cdot \frac{1}{2}=\frac{1}{16}$
• $\underset{x\to b}{lim}\frac{xsinb-bsinx}{x-b}=\underset{x\to b}{lim}\frac{xsinb-bsinb+bsinb-bsinx}{x-b}=\underset{x\to b}{lim}\frac{(x-b)sinb+b(sinb-sinx)}{x-b}=\underset{x\to b}{lim}\frac{(x-b)sinb}{x-b}+\underset{x\to b}{lim}\frac{b(sinb-sinx)}{x-b}=sinb+b\underset{x\to b}{lim}\frac{sinb-sinx}{x-b}=sinb+b\underset{x\to b}{lim}\frac{2cos\frac{b+x}{2}sin\frac{b-x}{2}}{x-b}=sinb+b\underset{x\to b}{lim}\frac{2cos\frac{b+x}{2}sin\frac{b-x}{2}}{-2\frac{b-x}{2}}=sinb-b\underset{x\to b}{lim}\frac{cos\frac{b+x}{2}sin\frac{b-x}{2}}{\frac{b-x}{2}}=sinb-b\underset{x\to b}{lim}cos\frac{b+x}{2}\cdot \underset{x\to b}{lim}\frac{sin\frac{b-x}{2}}{\frac{b-x}{2}}=sinb-bcosb\cdot 1=sinb-bcosb$
• $\underset{x\to \mathrm{\infty }}{lim}\frac{16x^5+12x+3}{4x^5+108x^3+67x}=\underset{x\to \mathrm{\infty }}{lim}\frac{x^5(16+\frac{12}{x^4}+\frac{3}{x^5})}{x^5(4+\frac{108}{x^2}+\frac{67}{x^4})}=\frac{16}{4}=4$
• $\underset{x\to \mathrm{\infty }}{lim}\frac{2^{5x+4}-5^{2x}}{2^{5x+3}+7^x}=\underset{x\to \mathrm{\infty }}{lim}\frac{2^{5x}\cdot 2^4-5^{2x}}{2^{5x}\cdot 2^3+7^x}=\underset{x\to \mathrm{\infty }}{lim}\frac{32^x\cdot 16-25^x}{32^x\cdot 8+7^x}=\underset{x\to \mathrm{\infty }}{lim}\frac{32^x(16-(\frac{25}{32}{)}^x)}{32^x(8+(\frac{7}{32}{)}^x)}=\frac{16}{8}=2$
• $\underset{x\to 0}{lim}\frac{4e^{3x}-5(2{)}^{9x}+1}{3x}=\underset{x\to 0}{lim}\frac{4e^{3x}-5(2^3{)}^{3x}-4+5}{3x}=\underset{x\to 0}{lim}\frac{4e^{3x}-4-(5(8{)}^{3x}-5)}{3x}=\underset{x\to 0}{lim}\frac{4e^{3x}-4}{3x}-\underset{x\to 0}{lim}\frac{5(8{)}^{3x}-5}{3x}=4\underset{x\to 0}{lim}\frac{e^{3x}-1}{3x}-5\underset{x\to 0}{lim}\frac{8^{3x}-1}{3x}=4-5ln8$
• $\underset{x\to 0}{lim}\frac{9\cdot 15^{3x}-3^{3x+2}}{27x}=\underset{x\to 0}{lim}\frac{9\cdot (3\cdot 5{)}^{3x}-3^{3x}\cdot 3^2}{27x}=\underset{x\to 0}{lim}\frac{9\cdot 3^{3x}\cdot 5^{3x}-3^{3x}\cdot 9}{27x}=\underset{x\to 0}{lim}\frac{9\cdot 3^{3x}\cdot (5^{3x}-1)}{27x}=\underset{x\to 0}{lim}\frac{9\cdot 3^{3x}\cdot 5^{3x}-3^{3x}\cdot 9}{27x}=\underset{x\to 0}{lim}\frac{9\cdot 3^{3x}\cdot (5^{3x}-1)}{9\cdot 3x}=3^{0}ln5=ln5$
• $\underset{x\to \mathrm{\infty }}{lim}\sqrt{9x^2+12x+7}+\sqrt{4x^2+8x+3}-5x=\underset{x\to \mathrm{\infty }}{lim}\sqrt{9x^2+12x+7}+\sqrt{4x^2+8x+3}-3x-2x=\underset{x\to \mathrm{\infty }}{lim}(\sqrt{9x^2+12x+7}-3x)+(\sqrt{4x^2+8x+3}-2x)=\underset{x\to \mathrm{\infty }}{lim}(\sqrt{9x^2+12x+7}-3x)+\underset{x\to \mathrm{\infty }}{lim}(\sqrt{4x^2+8x+3}-2x)=\underset{x\to \mathrm{\infty }}{lim}(\sqrt{9x^2+12x+7}-\sqrt{9x^2})+\underset{x\to \mathrm{\infty }}{lim}(\sqrt{4x^2+8x+3}-\sqrt{4x^2})=\frac{12-0}{2\sqrt{9}}+\frac{8-0}{2\sqrt{4}}=2+2=4$
• $\underset{x\to \mathrm{\infty }}{lim}(\frac{n+1}{n-3}{)}^{\frac{3n-9}{4}}=\underset{x\to \mathrm{\infty }}{lim}(\frac{n-3+4}{n-3}{)}^{\frac{3n-9}{4}}=\underset{x\to \mathrm{\infty }}{lim}(\frac{n-3}{n-3}+\frac{4}{n-3}{)}^{\frac{3(n-3)}{4}}=\underset{x\to \mathrm{\infty }}{lim}(1+\frac{1}{\frac{n-3}{4}}{)}^{\frac{3(n-3)}{4}}=(\underset{x\to \mathrm{\infty }}{lim}(1+\frac{1}{\frac{n-3}{4}}{)}^{\frac{n-3}{4}}{)}^3=e^3$
• $\underset{x\to \mathrm{\infty }}{lim}(\frac{n^2+4n+10}{n^2-4n+7}{)}^{\frac{2n^2-8n+14}{24n+21}}=\underset{x\to \mathrm{\infty }}{lim}(\frac{n^2+8n-4n+7+3}{n^2-4n+7}{)}^{\frac{2n^2-8n+14}{24n+9}}=\underset{x\to \mathrm{\infty }}{lim}(\frac{(n^2-4n+7)+(8n+3)}{n^2-4n+7}{)}^{\frac{2n^2-8n+14}{24n+9}}=\underset{x\to \mathrm{\infty }}{lim}(\frac{n^2-4n+7}{n^2-4n+7}+\frac{8n+3}{n^2-4n+7}{)}^{\frac{2n^2-8n+14}{24n+9}}=\underset{x\to \mathrm{\infty }}{lim}(1+\frac{1}{\frac{n^2-4n+7}{8n+3}}{)}^{\frac{2(n^2-4n+7)}{3(8n+3)}}=(\underset{x\to \mathrm{\infty }}{lim}(1+\frac{1}{\frac{n^2-4n+7}{8n+3}}{)}^{\frac{n^2-4n+7}{8n+3}}{)}^{\frac{2}{3}}=e^{\frac{2}{3}}=\sqrt[3]{e^2}$
• $\underset{x\to \mathrm{\infty }}{lim}(\frac{3n+8}{3n+7}{)}^{6n-5}=\underset{x\to \mathrm{\infty }}{lim}(\frac{3n+7+1}{3n+7}{)}^{6n-5}=\underset{x\to \mathrm{\infty }}{lim}(\frac{3n+7}{3n+7}+\frac{1}{3n+7}{)}^{6n-5}=\underset{x\to \mathrm{\infty }}{lim}(1+\frac{1}{3n+7}{)}^{6n-5}=\underset{x\to \mathrm{\infty }}{lim}(1+\frac{1}{3n+7}{)}^{\frac{(3n+7)(6n-5)}{3n+7}}=(\underset{x\to \mathrm{\infty }}{lim}(1+\frac{1}{3n+7}{)}^{3n+7}{)}^{\frac{6n-5}{3n+7}}=(\underset{x\to \mathrm{\infty }}{lim}(1+\frac{1}{3n+7}{)}^{3n+7}{)}^{\underset{x\to \mathrm{\infty }}{lim}\frac{6n-5}{3n+7}}=(\underset{x\to \mathrm{\infty }}{lim}(1+\frac{1}{3n+7}{)}^{3n+7}{)}^{\underset{x\to \mathrm{\infty }}{lim}\frac{n(6-\frac{5}{n})}{n(3+\frac{7}{n})}}=e^{\frac{6}{3}}=e^2$
• $\underset{x\to \mathrm{\infty }}{lim}(\frac{4n+3}{3n+1}{)}^{\frac{3n+1}{2n+5}}=\underset{x\to \mathrm{\infty }}{lim}(\frac{3n+1+n+2}{3n+1}{)}^{\frac{3n+1}{2n+5}}=\underset{x\to \mathrm{\infty }}{lim}(\frac{3n+1}{3n+1}+\frac{n+2}{3n+1}{)}^{\frac{3n+1}{2n+5}}=\underset{x\to \mathrm{\infty }}{lim}(1+\frac{n+2}{3n+1}{)}^{\frac{3n+1}{2n+5}}=\underset{x\to \mathrm{\infty }}{lim}(1+\frac{1}{\frac{3n+1}{n+2}}{)}^{\frac{3n+1}{2n+5}}=\underset{x\to \mathrm{\infty }}{lim}(1+\frac{1}{\frac{3n+1}{n+2}}{)}^{\frac{3n+1}{n+2}\cdot \frac{n+2}{2n+5}}=(\underset{x\to \mathrm{\infty }}{lim}(1+\frac{1}{\frac{3n+1}{n+2}}{)}^{\frac{3n+1}{n+2}}{)}^{\frac{n+2}{2n+5}}=(\underset{x\to \mathrm{\infty }}{lim}(1+\frac{1}{\frac{3n+1}{n+2}}{)}^{\frac{3n+1}{n+2}}{)}^{\underset{x\to \mathrm{\infty }}{lim}\frac{n+2}{2n+5}}=(\underset{x\to \mathrm{\infty }}{lim}(1+\frac{1}{\frac{3n+1}{n+2}}{)}^{\frac{3n+1}{n+2}}{)}^{\underset{x\to \mathrm{\infty }}{lim}\frac{n(1+\frac{2}{n})}{n(2+\frac{5}{n})}}=e^{\frac{1}{2}}=\sqrt{e}$
• $\underset{x\to \mathrm{\infty }}{lim}\frac{5n^2}{1+3+5+\dots +n}=\underset{x\to \mathrm{\infty }}{lim}\frac{5n^2}{n^2}=5$
• $\underset{x\to 5^{+}}{lim}\frac{x^2-x+20}{|x-5|}=\underset{x\to 5^{+}}{lim}\frac{(x-5)(x+4)}{(x-5)}=\underset{x\to 5^{+}}{lim}x+4=5+4=9$
• $\underset{x\to -6^{-}}{lim}\frac{|x+6|}{x^2+2x-24}=\underset{x\to -6^{-}}{lim}\frac{-(x+6)}{(x+6)(x-4)}=\underset{x\to -6^{-}}{lim}\frac{-1}{x-4}=\frac{-1}{-6-4}=\frac{-1}{-10}=\frac{1}{10}$
• $\underset{x\to 0}{lim}\frac{40^{2x}-8^{2x}-5^{2x}+1}{4x^2}=\underset{x\to 0}{lim}\frac{8^{2x}(5^{2x}-1)-(5^{2x}-1)}{4x^2}=\underset{x\to 0}{lim}\frac{(8^{2x}-1)(5^{2x}-1)}{4x^2}=\underset{x\to 0}{lim}\frac{(8^{2x}-1)(5^{2x}-1)}{2x\cdot 2x}=ln8\cdot ln5=ln40$
• $\underset{x\to 4}{lim}\frac{x^2-5x+4}{x^2-x-12}=\underset{x\to 4}{lim}\frac{2x-5}{2x-1}=\frac{2(4)-5}{2(4)-1}=\frac{3}{7}$

alternatif

• $\underset{x\to \mathrm{\infty }}{lim}\sqrt{9x^2+12x+7}+\sqrt{4x^2+8x+3}-5x$
  • $\underset{x\to \mathrm{\infty }}{lim}\sqrt{9x^2+12x+7}+\sqrt{4x^2+8x+3}-5x$
  • $\underset{x\to \mathrm{\infty }}{lim}\sqrt{(3x+2{)}^2}+\sqrt{(2x+2{)}^2}-5x$
  • $\underset{x\to \mathrm{\infty }}{lim}3x+2+2x+2-5x$
  • 4

1. tentukan hasil dari $\underset{x\to 0}{lim}(\frac{sinx}{x}{)}^{\frac{1}{x}}$!

Jawaban

$$\begin{array}{c}\underset{x\to 0}{lim}(\frac{sinx}{x}{)}^{\frac{1}{x}}\\ \underset{x\to 0}{lim}(1+\frac{sinx}{x}-1{)}^{\frac{1}{x}}\\ \underset{x\to 0}{lim}(1+\frac{sinx-x}{x}{)}^{\frac{1}{x}}\\ \underset{x\to 0}{lim}(1+\frac{sinx-x}{x}{)}^{(\frac{1}{\frac{sinx-x}{x}}\cdot \frac{sinx-x}{x}\cdot \frac{1}{x})}\\ (\underset{x\to 0}{lim}(1+\frac{sinx-x}{x}{)}^{\frac{1}{\frac{sinx-x}{x}}}{)}^{(\frac{sinx-x}{x}\cdot \frac{1}{x})}\\ e^{\underset{x\to 0}{lim}(\frac{sinx-x}{x}\cdot \frac{1}{x})}\\ e^{\underset{x\to 0}{lim}\frac{sinx-x}{x^2}}\\ \text{misalkan }p& =\underset{x\to 0}{lim}\frac{sinx-x}{x^2}\\ p& =\underset{x\to 0}{lim}\frac{sinx-x}{x^2}\\ \text{ubah x menjadi 3x}\\ p& =\underset{x\to 0}{lim}\frac{sin3x-3x}{(3x{)}^2}\\ p& =\underset{x\to 0}{lim}\frac{3sinx-4si{n}^{3}x-3x}{9x^2}\\ p& =\underset{x\to 0}{lim}\frac{3(sinx-x)-4si{n}^{3}x}{9x^2}\\ p& =\underset{x\to 0}{lim}\frac{3(sinx-x)}{9x^2}-\underset{x\to 0}{lim}\frac{4si{n}^{3}x}{9x^2}\\ p& =\frac{3}{9}\underset{x\to 0}{lim}\frac{sinx-x}{x^2}-\underset{x\to 0}{lim}\frac{4\cdot sinx\cdot si{n}^{2}x}{9x^2}\\ p& =\frac{1}{3}p-0\\ \frac{2}{3}p& =0\\ p& =0\\ \underset{x\to 0}{lim}\frac{sinx-x}{x^2}& =0\\ e^{\underset{x\to 0}{lim}\frac{sinx-x}{x^2}}\\ e^0\\ 1\\ \underset{x\to 0}{lim}(\frac{sinx}{x}{)}^{\frac{1}{x}}& =1\end{array}$$

1. tentukan hasil dari $\underset{x\to 0}{lim}(\frac{tanx}{x}{)}^{\frac{1}{x^2}}$!

Jawaban

$$\begin{array}{c}\underset{x\to 0}{lim}(\frac{tanx}{x}{)}^{\frac{1}{x^2}}\\ \underset{x\to 0}{lim}(1+\frac{tanx}{x}-1{)}^{\frac{1}{x^2}}\\ \underset{x\to 0}{lim}(1+\frac{tanx-x}{x}{)}^{\frac{1}{x^2}}\\ \underset{x\to 0}{lim}(1+\frac{tanx-x}{x}{)}^{(\frac{1}{\frac{tanx-x}{x}}\cdot \frac{tanx-x}{x}\cdot \frac{1}{x^2})}\\ (\underset{x\to 0}{lim}(1+\frac{tanx-x}{x}{)}^{\frac{1}{\frac{tanx-x}{x}}}{)}^{(\frac{tanx-x}{x}\cdot \frac{1}{x^2})}\\ e^{\underset{x\to 0}{lim}(\frac{tanx-x}{x}\cdot \frac{1}{x^2})}\\ e^{\underset{x\to 0}{lim}\frac{tanx-x}{x^3}}\\ \text{misalkan }p& =\underset{x\to 0}{lim}\frac{tanx-x}{x^3}\\ p& =\underset{x\to 0}{lim}\frac{tanx-x}{x^3}\\ p& =\underset{x\to 0}{lim}\frac{se{c}^{2}x-1}{3x^2}\text{(gunakan aturan L'Hôpital) }\\ p& =\underset{x\to 0}{lim}\frac{ta{n}^{2}x}{3x^2}\\ p& =\frac{1}{3}\\ \underset{x\to 0}{lim}\frac{tanx-x}{x^3}& =\frac{1}{3}\\ e^{\underset{x\to 0}{lim}\frac{tanx-x}{x^3}}\\ e^{\frac{1}{3}}\\ \sqrt[3]{e}\\ \underset{x\to 0}{lim}(\frac{tanx}{x}{)}^{\frac{1}{x^2}}& =\sqrt[3]{e}\end{array}$$

1. tentukan nilai a dan b dari $\underset{x\to 5}{lim}\frac{x^2+(a+b)x-10}{x^2+(a+6)x-15b}=\frac{7}{11}$!

Jawaban

$$\begin{array}{cc}\underset{x\to 5}{lim}\frac{x^2+(a+b)x-10}{x^2+(a+6)x-15b}& =\frac{7}{11}\\ x^2+(a+b)x-10& =0\\ (5{)}^2+(a+b)5-10& =0\\ 25+5a+5b-10& =0\\ 5a+5b& =-15\\ a+b& =-3\\ \underset{x\to 5}{lim}\frac{2x+(a+b)}{2x+(a+6)}& =\frac{7}{11}\\ \frac{2x+(a+b)}{2x+(a+6)}& =\frac{7}{11}\\ \frac{2(5)+(a+b)}{2(5)+(a+6)}& =\frac{7}{11}\\ \frac{10+(a+b)}{10+(a+6)}& =\frac{7}{11}\\ \frac{a+b+10}{a+16}& =\frac{7}{11}\\ 11(a+b+10)& =7(a+16)\\ 11a+11b+110& =7a+112\\ 4a+11b& =2\\ a+b& =-3\\ a& =-3-b\\ 4a+11b& =2\\ 4(-3-b)+11b& =2\\ -12-4b+11b& =2\\ 7b& =14\\ b& =2\\ a& =-3-b\\ a& =-3-2\\ a& =-5\end{array}$$