Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 185555 by yaslm last updated on 23/Jan/23

Answered by SEKRET last updated on 24/Jan/23

$$\mathbf{y}=\mathbf{a}\cdot \mathbf{sin}\left(\mathbf{t}\right)\mathbf{x}=\mathbf{a}\cdot \mathbf{cos}\left(\mathbf{t}\right){\mid }_{(\mathbf{x},\mathbf{y})\to 0}\to {\mid }_{\mathbf{a}\to 0}$$$${\mathbf{lim}}_{\mathbf{a}\to 0}\frac{\mathbf{a}\mathbf{sin}\left(\mathbf{t}\right)}{{\mathbf{a}}^2\cdot {\mathbf{sin}}^2\left(\mathbf{t}\right)+1}={\mathbf{lim}}_{\mathbf{a}\to 0}\frac{\mathrm{a}\cdot \mathbf{sin}\left(\mathbf{t}\right)}{{\mathbf{a}}^2\cdot {\mathbf{sin}}^2\left(\mathbf{t}\right)+{\mathbf{sin}}^2\left(\mathbf{t}\right)+{\mathbf{cos}}^2\left(\mathbf{t}\right)}$$$$={\mathbf{lim}}_{\mathbf{a}\to 0}\frac{1}{\mathbf{a}\cdot \mathbf{sin}\left(\mathbf{t}\right)+\frac{\mathbf{sin}\left(\mathbf{t}\right)}{\mathbf{a}}+\frac{{\mathbf{cos}}^2\left(\mathbf{t}\right)}{\mathbf{a}\cdot \mathbf{sin}\left(\mathbf{t}\right)}}=$$$$=\frac{1}{0+\mathrm{\infty }+\mathrm{\infty }}=\frac{1}{\mathrm{\infty }}=0$$

Commented by yaslm last updated on 24/Jan/23

nice, but I need delta , apselon

Answered by LEKOUMA last updated on 24/Jan/23

$$\underset{(x,y)\to (0,0)}{\lim }\frac{y}{x^2+1}=\frac{0}{0+1}=0$$$$\underset{(x,y)\to (0,0)}{\lim }\frac{y}{x^2+1}=0$$

Commented by yaslm last updated on 24/Jan/23

nice, but I need delta ,apselon

Answered by aba last updated on 29/Jan/23

$$\mathrm{\forall }ϵ>0\mid \frac{\mathrm{y}}{{\mathrm{x}}^2+1}\mid =\frac{\mid \mathrm{y}\mid }{{\mathrm{x}}^2+1}⩽\frac{\mid \mathrm{y}\mid }{1}=\sqrt{{\mathrm{y}}^2}=\sqrt{{(\mathrm{y}-0)}^2}⩽\sqrt{{(\mathrm{x}-0)}^2+{(\mathrm{y}-0)}^2}⩽\delta$$$$\mathrm{\forall }ϵ>0\mathrm{choose}\delta =ϵ>0,\mathrm{\forall }(\mathrm{x},\mathrm{y})\in {\mathbb{R}}^2:0<\sqrt{{(\mathrm{x}-0)}^2+{(\mathrm{y}-0)}^2}<\delta \Rightarrow \mid \frac{\mathrm{y}}{{\mathrm{x}}^2+1}\mid ⩽ϵ$$$$\iff \underset{(x,\mathrm{y})\to (0,0)}{\lim }\frac{\mathrm{y}}{{\mathrm{x}}^2+1}=0$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com