Soal-Soal Matematika/Limit

Rumus umum

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

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

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

Aturan L'Hôpital

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

contoh soal

1. tentukan nilai limit dari

• $\underset{x\to 3}{\lim }\frac{x^2-5x+6}{x^2-9}$
• $\underset{x\to p}{\lim }\frac{x^2-(3+p)x+3p}{x^2+(4-p)x-4p}$
• $\underset{x\to 9}{\lim }\frac{\sqrt{x}-3}{x^2-10x+9}$
• $\underset{x\to 8}{\lim }\frac{x^2-10x+16}{\sqrt[3]{x}-2}$
• $\underset{x\to 4}{\lim }\frac{(\sqrt{x}-2)(5x+4)}{x^2-5x+4}$
• $\underset{x\to 1}{\lim }\frac{(x-1{)}^2}{\sqrt[3]{x^2}-2\sqrt[3]{x}+1}$
• $\underset{x\to 0}{\lim }\frac{1-cosx}{8x^2}$
• $\underset{x\to b}{\lim }\frac{xsinb-bsinx}{x-b}$
• $\underset{x\to \mathrm{\infty }}{\lim }\frac{16x^5+12x+3}{4x^5+108x^3+67x}$
• $\underset{x\to \mathrm{\infty }}{\lim }\sqrt{9x^2+11x+2}-\sqrt{9x^2+5x+3}$
• $\underset{x\to \mathrm{\infty }}{\lim }(\frac{n+1}{n-3}{)}^{\frac{3n-9}{4}}$
• $\underset{x\to \mathrm{\infty }}{\lim }(\frac{n^2+4n+10}{n^2-4n+7}{)}^{\frac{2n^2-8n+14}{24n+9}}$
• $\underset{x\to \mathrm{\infty }}{\lim }(\frac{3n+8}{3n+7}{)}^{6n-5}$
• $\underset{x\to 5^{+}}{\lim }\frac{x^2-x+20}{|x-5|}$
• $\underset{x\to -6^{-}}{\lim }\frac{|x+6|}{x^2+2x-24}$
• $\underset{x\to 4}{\lim }\frac{x^2-5x+4}{x^2-x-12}$ (aturan L'Hôpital)

Jawab

• $\underset{x\to 3}{\lim }\frac{x^2-5x+6}{x^2-9}=\underset{x\to 3}{\lim }\frac{(x-3)(x-2)}{(x-3)(x+3)}=\underset{x\to 3}{\lim }\frac{x-2}{x+3}=\frac{3-2}{3+3}=\frac{1}{6}$
• $\underset{x\to p}{\lim }\frac{x^2-(3+p)x+3p}{x^2+(4-p)x-4p}=\underset{x\to p}{\lim }\frac{(x-3)(x-p)}{(x+4)(x-p)}=\underset{x\to p}{\lim }\frac{x-3}{x+4}=\frac{p-3}{p+4}$
• $\underset{x\to 9}{\lim }\frac{\sqrt{x}-3}{x^2-10x+9}=\underset{x\to 9}{\lim }\frac{(\sqrt{x}-3)(\sqrt{x}+3)}{(x-1)(x-9)(\sqrt{x}+3)}=\underset{x\to 9}{\lim }\frac{x-9}{(x-1)(x-9)(\sqrt{x}+3)}=\underset{x\to 9}{\lim }\frac{1}{(x-1)(\sqrt{x}+3)}=\frac{1}{(9-1)(\sqrt{9}+3)}=\frac{1}{48}$
• $\underset{x\to 8}{\lim }\frac{x^2-10x+16}{\sqrt[3]{x}-2}=\underset{x\to 8}{\lim }\frac{(x-8)(x-2)(\sqrt[3]{x^2}+2\sqrt[3]{x}+4)}{(\sqrt[3]{x}-2)(\sqrt[3]{x^2}+2\sqrt[3]{x}+4)}=\underset{x\to 8}{\lim }\frac{(x-8)(x-2)(\sqrt[3]{x^2}+2\sqrt[3]{x}+4)}{x-8}=\underset{x\to 8}{\lim }(x-2)(\sqrt[3]{x^2}+2\sqrt[3]{x}+4)=(8-2)(\sqrt[3]{8^2}+2\sqrt[3]{8}+4)=6(4+4+4)=72$
• $\underset{x\to 4}{\lim }\frac{(\sqrt{x}-2)(5x+4)}{x^2-5x+4}=\underset{x\to 4}{\lim }\frac{(\sqrt{x}-2)(5x+4)}{(x-4)(x-1)}=\underset{x\to 4}{\lim }\frac{(\sqrt{x}-2)(5x+4)}{(\sqrt{x}-2)(\sqrt{x}+2)(x-1)}=\underset{x\to 4}{\lim }\frac{5x+4}{(\sqrt{x}+2)(x-1)}=\frac{5(4)+4}{(\sqrt{4}+2)(4-1)}=2$
• $\underset{x\to 1}{\lim }\frac{(x-1{)}^2}{\sqrt[3]{x^2}-2\sqrt[3]{x}+1}=\underset{x\to 1}{\lim }\frac{({\sqrt[3]{x}}^3-1^3{)}^2}{\sqrt[3]{x^2}-2\sqrt[3]{x}+1}=\underset{x\to 1}{\lim }\frac{({\sqrt[3]{x}}^3-1^3{)}^2}{(\sqrt[3]{x}-1{)}^2}=\underset{x\to 1}{\lim }\frac{((\sqrt[3]{x}-1)(\sqrt[3]{x^2}+\sqrt[3]{x}+1){)}^2}{(\sqrt[3]{x}-1{)}^2}=\underset{x\to 1}{\lim }\frac{(\sqrt[3]{x}-1{)}^2(\sqrt[3]{x^2}+\sqrt[3]{x}+1{)}^2}{(\sqrt[3]{x}-1{)}^2}=\underset{x\to 1}{\lim }(\sqrt[3]{x^2}+\sqrt[3]{x}+1{)}^2=(\sqrt[3]{1^2}+\sqrt[3]{1}+1{)}^2=(1+1+1{)}^2=9$
• $\underset{x\to 0}{\lim }\frac{1-cosx}{8x^2}=\underset{x\to 0}{\lim }\frac{2si{n}^2\frac{x}{2}}{8x^2}=\frac{2sin\frac{x}{2}sin\frac{x}{2}}{8xx}=\frac{2}{8}\cdot \frac{1}{2}\cdot \frac{1}{2}=\frac{1}{16}$
• $\underset{x\to b}{\lim }\frac{xsinb-bsinx}{x-b}=\underset{x\to b}{\lim }\frac{xsinb-bsinb+bsinb-bsinx}{x-b}=\underset{x\to b}{\lim }\frac{(x-b)sinb+b(sinb-sinx)}{x-b}=\underset{x\to b}{\lim }\frac{(x-b)sinb}{x-b}+\underset{x\to b}{\lim }\frac{b(sinb-sinx)}{x-b}=sinb+b\underset{x\to b}{\lim }\frac{sinb-sinx}{x-b}=sinb+b\underset{x\to b}{\lim }\frac{2cos\frac{b+x}{2}sin\frac{b-x}{2}}{x-b}=sinb+b\underset{x\to b}{\lim }\frac{2cos\frac{b+x}{2}sin\frac{b-x}{2}}{-2\frac{b-x}{2}}=sinb-b\underset{x\to b}{\lim }\frac{cos\frac{b+x}{2}sin\frac{b-x}{2}}{\frac{b-x}{2}}=sinb-b\underset{x\to b}{\lim }cos\frac{b+x}{2}\cdot \underset{x\to b}{\lim }\frac{sin\frac{b-x}{2}}{\frac{b-x}{2}}=sinb-bcosb\cdot 1=sinb-bcosb$
• $\underset{x\to \mathrm{\infty }}{\lim }\frac{16x^5+12x+3}{4x^5+108x^3+67x}=\underset{x\to \mathrm{\infty }}{\lim }\frac{x^5(16+\frac{12}{x^4}+\frac{3}{x^5})}{x^5(4+\frac{108}{x^2}+\frac{67}{x^4})}=\frac{16}{4}=4$
• $\underset{x\to \mathrm{\infty }}{\lim }\sqrt{9x^2+11x+2}-\sqrt{9x^2+5x+3}=\frac{11-5}{2\sqrt{9}}=\frac{6}{2\cdot 3}=1$
• $\underset{x\to \mathrm{\infty }}{\lim }(\frac{n+1}{n-3}{)}^{\frac{3n-9}{4}}=\underset{x\to \mathrm{\infty }}{\lim }(\frac{n-3+4}{n-3}{)}^{\frac{3n-9}{4}}=\underset{x\to \mathrm{\infty }}{\lim }(\frac{n-3}{n-3}+\frac{4}{n-3}{)}^{\frac{3(n-3)}{4}}=\underset{x\to \mathrm{\infty }}{\lim }(1+\frac{1}{\frac{n-3}{4}}{)}^{\frac{3(n-3)}{4}}=(\underset{x\to \mathrm{\infty }}{\lim }(1+\frac{1}{\frac{n-3}{4}}{)}^{\frac{n-3}{4}}{)}^3=e^3$
• $\underset{x\to \mathrm{\infty }}{\lim }(\frac{n^2+4n+10}{n^2-4n+7}{)}^{\frac{2n^2-8n+14}{24n+21}}=\underset{x\to \mathrm{\infty }}{\lim }(\frac{n^2+8n-4n+7+3}{n^2-4n+7}{)}^{\frac{2n^2-8n+14}{24n+9}}=\underset{x\to \mathrm{\infty }}{\lim }(\frac{(n^2-4n+7)+(8n+3)}{n^2-4n+7}{)}^{\frac{2n^2-8n+14}{24n+9}}=\underset{x\to \mathrm{\infty }}{\lim }(\frac{n^2-4n+7}{n^2-4n+7}+\frac{8n+3}{n^2-4n+7}{)}^{\frac{2n^2-8n+14}{24n+9}}=\underset{x\to \mathrm{\infty }}{\lim }(1+\frac{1}{\frac{n^2-4n+7}{8n+3}}{)}^{\frac{2(n^2-4n+7)}{3(8n+3)}}=(\underset{x\to \mathrm{\infty }}{\lim }(1+\frac{1}{\frac{n^2-4n+7}{8n+3}}{)}^{\frac{n^2-4n+7}{8n+3}}{)}^{\frac{2}{3}}=e^{\frac{2}{3}}=\sqrt[3]{e^2}$
• $\underset{x\to \mathrm{\infty }}{\lim }(\frac{3n+8}{3n+7}{)}^{6n-5}=\underset{x\to \mathrm{\infty }}{\lim }(\frac{3n+7+1}{3n+7}{)}^{6n-5}=\underset{x\to \mathrm{\infty }}{\lim }(\frac{3n+7}{3n+7}+\frac{1}{3n+7}{)}^{6n-5}=\underset{x\to \mathrm{\infty }}{\lim }(1+\frac{1}{3n+7}{)}^{6n-5}=\underset{x\to \mathrm{\infty }}{\lim }(1+\frac{1}{3n+7}{)}^{\frac{(3n+7)(6n-5)}{3n+7}}=(\underset{x\to \mathrm{\infty }}{\lim }(1+\frac{1}{3n+7}{)}^{3n+7}{)}^{\frac{6n-5}{3n+7}}=(\underset{x\to \mathrm{\infty }}{\lim }(1+\frac{1}{3n+7}{)}^{3n+7}{)}^{\underset{x\to \mathrm{\infty }}{\lim }\frac{6n-5}{3n+7}}=(\underset{x\to \mathrm{\infty }}{\lim }(1+\frac{1}{3n+7}{)}^{3n+7}{)}^{\underset{x\to \mathrm{\infty }}{\lim }\frac{n(6-\frac{5}{n})}{n(3+\frac{7}{n})}}=e^{\frac{6}{3}}=e^2$
• $\underset{x\to 5^{+}}{\lim }\frac{x^2-x+20}{|x-5|}=\underset{x\to 5^{+}}{\lim }\frac{(x-5)(x+4)}{(x-5)}=\underset{x\to 5^{+}}{\lim }x+4=5+4=9$
• $\underset{x\to -6^{-}}{\lim }\frac{|x+6|}{x^2+2x-24}=\underset{x\to -6^{-}}{\lim }\frac{-(x+6)}{(x+6)(x-4)}=\underset{x\to -6^{-}}{\lim }\frac{-1}{x-4}=\frac{-1}{-6-4}=\frac{-1}{-10}=\frac{1}{10}$
• $\underset{x\to 4}{\lim }\frac{x^2-5x+4}{x^2-x-12}=\underset{x\to 4}{\lim }\frac{2x-5}{2x-1}=\frac{2(4)-5}{2(4)-1}=\frac{3}{7}$

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

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