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Question Number 139432 by normabaru last updated on 27/Apr/21

Answered by Jme Eduardo last updated on 27/Apr/21

(a)  if  x=t^2   and  y=t    lim_(t→0)  [((2t^4 )/(t^4 +t^4 ))]= 1    if  x=t  and  y=t^2     lim [((2t^4 )/(t^2 +t^8 ))]= lim_(t→0)  [((2t)/(1+t^6 ))]= 0    the lim_((x;y)→(0;0))    f(x;y) not exist, because if we approach parametric curves the function has  different  limits

$$\left({a}\right)\:\:{if}\:\:{x}={t}^{\mathrm{2}} \:\:{and}\:\:{y}={t} \\ $$$$ \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\frac{\mathrm{2}{t}^{\mathrm{4}} }{{t}^{\mathrm{4}} +{t}^{\mathrm{4}} }\right]=\:\mathrm{1} \\ $$$$ \\ $$$${if}\:\:{x}={t}\:\:{and}\:\:{y}={t}^{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{lim}\:\left[\frac{\mathrm{2}{t}^{\mathrm{4}} }{{t}^{\mathrm{2}} +{t}^{\mathrm{8}} }\right]=\:\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{6}} }\right]=\:\mathrm{0} \\ $$$$ \\ $$$${the}\:\underset{\left({x};{y}\right)\rightarrow\left(\mathrm{0};\mathrm{0}\right)} {\mathrm{lim}}\:\:\:{f}\left({x};{y}\right)\:{not}\:{exist},\:{because}\:{if}\:{we}\:{approach}\:{parametric}\:{curves}\:{the}\:{function}\:{has}\:\:{different}\:\:{limits} \\ $$

Answered by Jme Eduardo last updated on 27/Apr/21

(a)  if  x=t^2   and  y=t    lim_(t→0)  [((2t^4 )/(t^4 +t^4 ))]= 1    if  x=t  and  y=t^2     lim [((2t^4 )/(t^2 +t^8 ))]= lim_(t→0)  [((2t)/(1+t^6 ))]= 0    the lim_((x;y)→(0;0))    f(x;y) not exist, because if we approach parametric curves the function has  different  limits

$$\left({a}\right)\:\:{if}\:\:{x}={t}^{\mathrm{2}} \:\:{and}\:\:{y}={t} \\ $$$$ \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\frac{\mathrm{2}{t}^{\mathrm{4}} }{{t}^{\mathrm{4}} +{t}^{\mathrm{4}} }\right]=\:\mathrm{1} \\ $$$$ \\ $$$${if}\:\:{x}={t}\:\:{and}\:\:{y}={t}^{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{lim}\:\left[\frac{\mathrm{2}{t}^{\mathrm{4}} }{{t}^{\mathrm{2}} +{t}^{\mathrm{8}} }\right]=\:\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{6}} }\right]=\:\mathrm{0} \\ $$$$ \\ $$$${the}\:\underset{\left({x};{y}\right)\rightarrow\left(\mathrm{0};\mathrm{0}\right)} {\mathrm{lim}}\:\:\:{f}\left({x};{y}\right)\:{not}\:{exist},\:{because}\:{if}\:{we}\:{approach}\:{parametric}\:{curves}\:{the}\:{function}\:{has}\:\:{different}\:\:{limits} \\ $$

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