Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

Question Number 2265 by B744237509 last updated on 12/Nov/15

lim_(x→0) ((3x+sin2x)/(2x+sin3x))=?

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{3}{x}+{sin}\mathrm{2}{x}}{\mathrm{2}{x}+{sin}\mathrm{3}{x}}=? \\ $$$$ \\ $$$$ \\ $$

Answered by Filup last updated on 12/Nov/15

=lim_(x→0) (((3x)/(2x+sin(3x)))+((sin(2x))/(2x+sin(3x))))  Use L′Ho^� pital′s law  =lim_(x→0) ((3/(2+3cos(3x)))+((2cos(2x))/(2+3cos(3x))))  cos(0)=1  =(3/(2+3cos(3×0)))+((2cos(2×0))/(2+3cos(3×0)))  =(3/(2+3))+(2/(2+3))  =(3/5)+(2/5)  =1  ∴lim_(x→0) ((3x+sin(2x))/(2x+sin(3x)))=1

$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{3}{x}}{\mathrm{2}{x}+\mathrm{sin}\left(\mathrm{3}{x}\right)}+\frac{\mathrm{sin}\left(\mathrm{2}{x}\right)}{\mathrm{2}{x}+\mathrm{sin}\left(\mathrm{3}{x}\right)}\right) \\ $$$${Use}\:{L}'{H}\hat {{o}pital}'{s}\:{law} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{3}}{\mathrm{2}+\mathrm{3cos}\left(\mathrm{3}{x}\right)}+\frac{\mathrm{2cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}+\mathrm{3cos}\left(\mathrm{3}{x}\right)}\right) \\ $$$$\mathrm{cos}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}+\mathrm{3cos}\left(\mathrm{3}×\mathrm{0}\right)}+\frac{\mathrm{2cos}\left(\mathrm{2}×\mathrm{0}\right)}{\mathrm{2}+\mathrm{3cos}\left(\mathrm{3}×\mathrm{0}\right)} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}+\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{2}+\mathrm{3}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{5}}+\frac{\mathrm{2}}{\mathrm{5}} \\ $$$$=\mathrm{1} \\ $$$$\therefore\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{3}{x}+\mathrm{sin}\left(\mathrm{2}{x}\right)}{\mathrm{2}{x}+\mathrm{sin}\left(\mathrm{3}{x}\right)}=\mathrm{1} \\ $$

Commented by Filup last updated on 12/Nov/15

Expanding the fraction into two parts  isnt nessisary. You can start on line 2

$$\mathrm{Expanding}\:\mathrm{the}\:\mathrm{fraction}\:\mathrm{into}\:\mathrm{two}\:\mathrm{parts} \\ $$$$\mathrm{isnt}\:\mathrm{nessisary}.\:\mathrm{You}\:\mathrm{can}\:\mathrm{start}\:\mathrm{on}\:\mathrm{line}\:\mathrm{2} \\ $$

Answered by Rasheed Soomro last updated on 12/Nov/15

A way which doesn′t use derivative.  lim_(x→0) ((3x+sin2x)/(2x+sin3x))  =lim_(x→0) ((((3x+sin2x)/(2x))×2x)/(((2x+sin3x)/(3x))×3x))  =(2/3)lim_(x→0) ((((3x)/(2x))+((sin 2x)/(2x)))/(((2x)/(3x))+((sin 3x)/(3x))))  =(2/3)lim_(x→0) (((3/2)+((sin 2x)/(2x)))/((2/3)+((sin 3x)/(3x))))  =(2/3)×(((3/2)+lim_(2x→0) ((sin 2x)/(2x)))/((2/3)+lim_(3x→0) ((sin 3x)/(3x))))     [x→0 { ((2x→0)),((3x→0)) :}  ]  =(2/3)×(((3/2)+1)/((2/3)+1))                      [lim_(x→0) ((sin x)/x)=1]   =(2/3)×(5/2)×(3/5)=1

$${A}\:{way}\:{which}\:{doesn}'{t}\:{use}\:{derivative}. \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{3}{x}+{sin}\mathrm{2}{x}}{\mathrm{2}{x}+{sin}\mathrm{3}{x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{3}{x}+{sin}\mathrm{2}{x}}{\mathrm{2}{x}}×\mathrm{2}{x}}{\frac{\mathrm{2}{x}+{sin}\mathrm{3}{x}}{\mathrm{3}{x}}×\mathrm{3}{x}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\frac{\mathrm{3}{x}}{\mathrm{2}{x}}+\frac{{sin}\:\mathrm{2}{x}}{\mathrm{2}{x}}}{\frac{\mathrm{2}{x}}{\mathrm{3}{x}}+\frac{{sin}\:\mathrm{3}{x}}{\mathrm{3}{x}}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\frac{\mathrm{3}}{\mathrm{2}}+\frac{{sin}\:\mathrm{2}{x}}{\mathrm{2}{x}}}{\frac{\mathrm{2}}{\mathrm{3}}+\frac{{sin}\:\mathrm{3}{x}}{\mathrm{3}{x}}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}×\frac{\frac{\mathrm{3}}{\mathrm{2}}+\underset{\mathrm{2}{x}\rightarrow\mathrm{0}} {{lim}}\frac{{sin}\:\mathrm{2}{x}}{\mathrm{2}{x}}}{\frac{\mathrm{2}}{\mathrm{3}}+\underset{\mathrm{3}{x}\rightarrow\mathrm{0}} {{lim}}\frac{{sin}\:\mathrm{3}{x}}{\mathrm{3}{x}}}\:\:\:\:\:\left[{x}\rightarrow\mathrm{0\begin{cases}{\mathrm{2}{x}\rightarrow\mathrm{0}}\\{\mathrm{3}{x}\rightarrow\mathrm{0}}\end{cases}}\:\:\right] \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}×\frac{\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{1}}{\frac{\mathrm{2}}{\mathrm{3}}+\mathrm{1}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{sin}\:{x}}{{x}}=\mathrm{1}\right] \\ $$$$\:=\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{5}}{\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{5}}=\mathrm{1} \\ $$

Commented by Filup last updated on 12/Nov/15

I knew there was a way but i didnt know  how to do it. Nice!

$$\mathrm{I}\:\mathrm{knew}\:\mathrm{there}\:\mathrm{was}\:\mathrm{a}\:\mathrm{way}\:\mathrm{but}\:\mathrm{i}\:\mathrm{didnt}\:\mathrm{know} \\ $$$$\mathrm{how}\:\mathrm{to}\:\mathrm{do}\:\mathrm{it}.\:\mathrm{Nice}! \\ $$

Commented by B744237509 last updated on 12/Nov/15

thanx

$${thanx} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com