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 161533 by mathlove last updated on 19/Dec/21

Commented by cortano last updated on 19/Dec/21

$$\underset{x\to 0}{\lim }\frac{{\left(\sqrt{1+x^2+x}\right)}^{2020}-{\left(\sqrt{1+x^2-x}\right)}^{2020}}{x}$$$$=\underset{x\to 0}{\lim }\frac{(1+1010(x^2+x))-(1+1010(x^2-x))}{x}$$$$=\underset{x\to 0}{\lim }\frac{2020x}{x}=2020$$

Answered by mathmax by abdo last updated on 19/Dec/21

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{{(1+{\mathrm{x}}^2+\mathrm{x})}^{1010}-{(1+{\mathrm{x}}^2-\mathrm{x})}^{1010}}{\mathrm{x}}\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)\sim \frac{1+1010({\mathrm{x}}^2+\mathrm{x})-(1+1010({\mathrm{x}}^2-\mathrm{x}))}{\mathrm{x}}$$$$\mathrm{f}\left(\mathrm{x}\right)\sim \frac{{1010\mathrm{x}}^2+1010\mathrm{x}-{1010\mathrm{x}}^2+1010\mathrm{x}}{\mathrm{x}}\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)\sim \frac{2020\mathrm{x}}{\mathrm{x}}\Rightarrow {\lim }_{\mathrm{x}\to 0}\mathrm{f}\left(\mathrm{x}\right)=2020$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com