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 30502 by abdo imad last updated on 22/Feb/18

$$find{lim}_{x\to 0}{(sinx+cosx)}^{\frac{1}{x}}.$$

Answered by sma3l2996 last updated on 24/Feb/18

$${(sinx+cosx)}^{1/x}=e^{\frac{ln(sinx+cosx)}{x}}$$$$sin\left(x\right)\underset{0}{\sim }x-\frac{x^3}{3!}andcos\left(x\right)\underset{0}{\sim }1-\frac{x^2}{2}$$$$sosinx+cosx\underset{0}{\sim }1+x-\frac{x^2}{2}-\frac{x^3}{6}$$$$ln(cosx+sinx)\underset{0}{\sim }ln(1+x-x^2/2-x^3/6)\underset{0}{\sim }x-x^2/2-x^3/6$$$$\frac{ln(sinx+cosx)}{x}\underset{0}{\sim }1$$$$so\underset{x\to 0}{lim}{(sinx+cosx)}^{1/x}=e$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com