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Question Number 77309 by TawaTawa last updated on 05/Jan/20

Commented by mr W last updated on 05/Jan/20

$$=\mathrm{\infty }$$

Commented by mathmax by abdo last updated on 05/Jan/20

$$letx=rcos\theta andy=rsin\theta \Rightarrow {lim}_{(x,y)\to (0,0)}={lim}_{r\to 0}\frac{r^{2}cos\theta sin\theta +2}{r^2}$$$$={lim}_{r\to 0}\frac{2}{r^2}=+\mathrm{\infty }$$

Commented by TawaTawa last updated on 05/Jan/20

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by mathmax by abdo last updated on 05/Jan/20

$$youarewelcome$$

Answered by MJS last updated on 05/Jan/20

$$\mathrm{case}1:x\to 0^{+}\wedge y\to 0^{+}\vee x\to 0^{-}\wedge y\to 0^{-}$$$$\Rightarrow y=x$$$$\underset{x\to 0}{\lim }\frac{x^2+2}{2x^2}=\underset{x\to 0}{\lim }(\frac{1}{x^2}+\frac{1}{2})=+\mathrm{\infty }$$$$\mathrm{case}2:x\to 0^{+}\wedge y\to 0^{-}\vee x\to 0^{-}\wedge y\to 0^{+}$$$$\Rightarrow y=-x$$$$\underset{x\to 0}{\lim }\frac{2-x^2}{2x^2}=\underset{x\to 0}{\lim }(\frac{1}{x^2}-\frac{1}{2})=+\mathrm{\infty }$$

Commented by TawaTawa last updated on 05/Jan/20

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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