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 187598 by horsebrand11 last updated on 19/Feb/23

$$\underset{x\to 0^{+}}{\lim }\frac{6}{x}-\sqrt{\frac{36}{x^2}+\frac{4}{x}+9}=?$$

Answered by cortano12 last updated on 19/Feb/23

$$L=\underset{x\to 0^{+}}{\lim }\frac{6}{x}-\frac{1}{x}\sqrt{36+4x+9x^2}$$$$L=\underset{x\to 0^{+}}{\lim }\frac{6-\sqrt{9x^2+4x+36}}{x}$$$$L=\underset{x\to 0^{+}}{\lim }\frac{-x(9x+4)}{x(6+\sqrt{9x^2+4x+36})}$$$$L=-\frac{1}{3}$$

Answered by a.lgnaoui last updated on 19/Feb/23

$$\mathrm{L}=\frac{6}{x}-\sqrt{{\left(\frac{6}{x}\right)}^2+\frac{2}{3}\left(\frac{6}{x}\right)+9}$$$$posons\mathrm{X}=\frac{6}{x}\Rightarrow \mathrm{X}-\sqrt{{\mathrm{X}}^2+\frac{2}{3}\mathrm{X}+9}$$$$\mathrm{X}-\sqrt{{(\mathrm{X}+\frac{1}{3})}^2+\frac{80}{9}}=\mathrm{X}-(\mathrm{X}+\frac{1}{3})\sqrt{1+\frac{80}{9{(\mathrm{X}+\frac{1}{3})}^2}}$$$$x\to 0^{0+}\Rightarrow \mathrm{X}\to +\mathrm{\infty }\mathrm{et}\sqrt{1+\frac{80}{9{(\mathrm{x}+\frac{1}{3})}^2}}\to 1$$$$$$$${\lim }_{x\to 0^{+}}\frac{6}{x}-\sqrt{\frac{36}{x^2}+\frac{4}{x}+9}=(-\frac{1}{3})$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com