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 176035 by cortano1 last updated on 11/Sep/22

$$\underset{x\to 0}{\lim }\frac{\mathrm{x}\sqrt{1+{\mathrm{x}}^2}-\sin \mathrm{x}}{{(1+{\mathrm{x}}^2)}^{\mathrm{x}}-\sqrt{1+{\mathrm{x}}^3}}=$$

Answered by Ar Brandon last updated on 11/Sep/22

$$\mathcal{L}=\underset{x\to 0}{\lim }\frac{x\sqrt{1+x^2}-\sin x}{{(1+x^2)}^x-\sqrt{1+x^3}}=\underset{x\to 0}{\lim }\frac{x(1+\frac{x^2}{2})-(x-\frac{x^3}{3!})}{(1+x^3)-(1+\frac{x^3}{2})}$$$$=\underset{x\to 0}{\lim }\frac{x+\frac{x^3}{2}-(x-\frac{x^3}{6})}{\frac{x^3}{2}}=\underset{x\to 0}{\lim }\frac{\frac{x^3}{2}+\frac{x^3}{6}}{\frac{x^3}{2}}=\frac{\left(4/6\right)}{1/2}=\frac{4}{3}$$

Commented by cortano1 last updated on 11/Sep/22

$$\mathrm{Yes}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com