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 120686 by bobhans last updated on 01/Nov/20

$$\underset{x\to 0}{\lim }\frac{\sin \mathrm{x}-\tan \mathrm{x}}{{\mathrm{x}}^3}?$$

Answered by john santu last updated on 02/Nov/20

$$standartway$$$$\underset{x\to 0}{\lim }\frac{\sin x-\tan x}{x^3}=\underset{x\to 0}{\lim }\frac{\tan x(\cos x-1)}{x^3}$$$$=\underset{x\to 0}{\lim }\frac{\tan x(-2\sin {}^2\left(\frac{1}{2}x\right))}{x^3}=-0.5$$

Answered by mathmax by abdo last updated on 02/Nov/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{sinx}-\mathrm{tanx}}{{\mathrm{x}}^3}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{sinx}-\frac{\mathrm{sinx}}{\mathrm{cosx}}}{{\mathrm{x}}^3}=\frac{\mathrm{sinxcosx}-\mathrm{sinx}}{{\mathrm{x}}^3\mathrm{cosx}}$$$$=\frac{\mathrm{sinx}}{\mathrm{x}}.\frac{\mathrm{cosx}-1}{{\mathrm{x}}^2\mathrm{cosx}}\mathrm{we}\mathrm{have}\frac{\mathrm{sinx}}{\mathrm{x}}\sim 1\mathrm{and}\mathrm{cosx}-1\sim -\frac{{\mathrm{x}}^2}{2}\Rightarrow$$$$\frac{\mathrm{cosx}}{{\mathrm{x}}^2\mathrm{cosx}}\sim -\frac{1}{2}\Rightarrow {\lim }_{\mathrm{x}\to 0}\mathrm{f}\left(\mathrm{x}\right)=-\frac{1}{2}$$

Answered by malwaan last updated on 02/Nov/20

$$\underset{x\to 0}{lim}\frac{(x-\frac{x^3}{3!}+..)-(x+\frac{x^3}{3}+..)}{x^3}$$$$=\underset{x\to 0}{lim}(-\frac{1}{6}-\frac{1}{3})=-\frac{1}{2}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com