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 193863 by horsebrand11 last updated on 21/Jun/23

$$\underset{x\to 0}{\lim }\frac{1-\frac{1}{2}{\mathrm{x}}^2-\cos \left(\frac{\mathrm{x}}{1-{\mathrm{x}}^2}\right)}{{\mathrm{x}}^4}=?$$

Answered by MM42 last updated on 21/Jun/23

$$u\to 0\Rightarrow 1-cosu\sim \frac{1}{2}{u}^2$$$$\Rightarrow {lim}_{x\to 0}\frac{1-cos\left(\frac{x}{1-x^2}\right)-\frac{1}{2}{x}^2}{x^4}$$$$={lim}_{x\to 0}\frac{\frac{1}{2}{\left(\frac{x}{1-x^2}\right)}^2-\frac{1}{2}{x}^2}{x^4}$$$$={lim}_{x\to 0}\frac{\frac{1}{2}(\frac{1}{{(1-x^2)}^2}-1)}{x^2}$$$$={lim}_{x\to 0}\frac{2x^2-x^4}{2x^2{(1-x^2)}^2}=$$$$={lim}_{x\to 0}\frac{2-x^2}{2{(1-x^2)}^2}=1✓$$$$$$

Commented by cortano12 last updated on 22/Jun/23

$$1=\mathrm{wrong}$$

Answered by cortano12 last updated on 22/Jun/23

$$\mathrm{answer}\mathrm{is}\frac{23}{24}$$

Commented by MM42 last updated on 22/Jun/23

$$why?$$$$pleaseprovideyoursolution$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com