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 153346 by mathlove last updated on 06/Sep/21

Answered by liberty last updated on 06/Sep/21

$$\underset{x\to 2}{\lim }\frac{f\left(x\right)-4}{x-2}=15\to \{\begin{array}{l}f\left(2\right)=4\\ f{}^{\prime }\left(2\right)=15\end{array}$$$$\underset{x\to 2}{\lim }\frac{2x^2-f\left(x\right)}{x-2}=\mathrm{\infty }$$

Answered by Mathspace last updated on 06/Sep/21

$$bylhospitalweget{lim}_{x\to 2}{f}^{{}^{\prime }}\left(x\right)=15$$$$\Rightarrow {lim}_{x\to 2}\frac{2x^2-f\left(x\right)}{x-2}={lim}_{x\to 2}4x-f^{{}^{\prime }}\left(x\right)$$$$=8-f^{{}^{\prime }}\left(2\right)=8-15=-7$$

Commented by liberty last updated on 07/Sep/21

$$wrong.$$$$\underset{x\to 2}{\lim }\frac{2x^2-f\left(x\right)}{x-2}\ne \frac{0}{0}$$$$L^{\prime }Hopitalnotavailable$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com