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 72462 by 20190927 last updated on 29/Oct/19

$$\underset{x\to 0}{\lim }\frac{1-\sqrt{1+4\mathrm{x}}\cos \left({\mathrm{x}}^2\right)}{{\mathrm{x}}^3\arctan \left({\mathrm{x}}^5\right)}$$

Commented by kaivan.ahmadi last updated on 29/Oct/19

$${lim}_{x\to 0}\frac{1-\sqrt{1+4x}(1-\frac{x^4}{2})}{x^8}={lim}_{x\to 0}\frac{2-2\sqrt{1+4x}+x^4\sqrt{1+4x}}{x^8}=$$$${lim}_{x\to 0}\frac{2}{x^8}=+\mathrm{\infty }$$

Commented by mathmax by abdo last updated on 29/Oct/19

$$wehave{(1+u)}^{\alpha }=1+\alpha u+\frac{\alpha (\alpha -1)}{2}{u}^2+o\left(u^2\right)\Rightarrow$$$$\sqrt{1+4x}={(1+4x)}^{\frac{1}{2}}=1+2x+\frac{1}{2}\left(\frac{1}{2}\right)(-\frac{1}{2})\left(16x^2\right)+o\left(x^2\right)$$$$=1+2x-2x^2+o\left(x^2\right)andcos\left(x^2\right)=1-\frac{x^4}{2}+o\left(x^4\right)\Rightarrow$$$$\sqrt{1+4x}cos\left(x^2\right)\sim (1+2x-2x^2)(1-\frac{x^4}{2})$$$$=1+2x-2x^2-\frac{x^4}{2}-x^3+x^6\Rightarrow 1-\sqrt{1+4x}cos\left(x^2\right)\sim -2x+2x^2+\frac{x^4}{2}+x^3+x^6$$$$also{x}^{3}arcan\left(x^5\right)\sim x^8\Rightarrow \frac{1-\sqrt{1+4x}cos\left(x^2\right)}{x^{3}arctan\left(x^5\right)}\sim \frac{-2+2x+\frac{x^3}{2}+x^2+x^5}{x^7}\Rightarrow$$$${lim}_{x\to 0}\frac{1-\sqrt{1+4x}cos\left(x^2\right)}{x^{3}arctan\left(x^5\right)}=\mathrm{\infty }$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com