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 104217 by bemath last updated on 20/Jul/20

$$\underset{x\to 0}{\lim }\frac{\tan {(x+3)}^2-\tan \left(9\right)}{x}=?$$

Answered by bemath last updated on 20/Jul/20

$$\underset{x\to 0}{\lim }\frac{2{(x+3)}^1\sec {}^2{(x+3)}^2}{1}=$$$$2.3.\sec {}^2\left(9\right)=6\sec {}^2\left(9\right)★$$

Answered by Dwaipayan Shikari last updated on 20/Jul/20

$$\underset{x\to 0}{\lim }\frac{\frac{\sin {(\mathrm{x}+3)}^2}{\cos {(\mathrm{x}+3)}^2}-\frac{\sin 9}{\cos 9}}{\mathrm{x}}=\underset{x\to 0}{\lim }\frac{\sin ({(\mathrm{x}+3)}^2-9)}{\cos {(\mathrm{x}+3)}^2\cos 9}.\frac{1}{\mathrm{x}}\left(\sin ({\mathrm{x}}^2+6\mathrm{x})\right)={\mathrm{x}}^2+6\mathrm{x}$$$$\underset{x\to 0}{\lim }\frac{\sin ({\mathrm{x}}^2+6\mathrm{x})}{\cos {(\mathrm{x}+3)}^2\cos 9}.\frac{1}{\mathrm{x}}\underset{x\to 0}{\lim }\frac{{\mathrm{x}}^2+6\mathrm{x}}{{\cos }^{2}9}.\frac{1}{\mathrm{x}}=6{\sec }^{2}9★$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com