O limite (x,y) =>(2,4) ( x + 4) / ( x²y - 3y +4x² -12) , vale :
a) 1
b) -1
c) 0
d) -8
e) 12
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Soluções para a tarefa
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1
Boa tarde Roger!
Para resolver esse limite,basta substituir os valores,pois o mesmo não apresenta indeterminação é só fazer a substituição.
![\displaystyle \lim_{(x,y) \to (2,4)} \frac{y+4}{ x^{2} y-3y+4 x^{2} -12} \\\\\\\\\\\
\displaystyle \lim_{(x,y) \to (2,4)} \frac{4+4}{ (2)^{2} .4-3(4)+4.(2)^{2} -12} \\\\\\\\\\\
\displaystyle \lim_{(x,y) \to (2,4)} \frac{4+4}{ 4.4-3(4)+4.(4) -12} \\\\\\\\\\\
\displaystyle \lim_{(x,y) \to (2,4)} \frac{4+4}{ 16-12+16 -12} \\\\\\\\\\\
\displaystyle \lim_{(x,y) \to (2,4)} \frac{8}{ 32-24} \\\\\\\\\\\
\displaystyle \lim_{(x,y) \to (2,4)} \frac{8}{8}](https://tex.z-dn.net/?f=%5Cdisplaystyle+%5Clim_%7B%28x%2Cy%29+%5Cto+%282%2C4%29%7D++%5Cfrac%7By%2B4%7D%7B+x%5E%7B2%7D+y-3y%2B4+x%5E%7B2%7D+-12%7D+%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%0A%0A%5Cdisplaystyle+%5Clim_%7B%28x%2Cy%29+%5Cto+%282%2C4%29%7D++%5Cfrac%7B4%2B4%7D%7B+%282%29%5E%7B2%7D+.4-3%284%29%2B4.%282%29%5E%7B2%7D++-12%7D+%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%0A%0A%0A%5Cdisplaystyle+%5Clim_%7B%28x%2Cy%29+%5Cto+%282%2C4%29%7D++%5Cfrac%7B4%2B4%7D%7B+4.4-3%284%29%2B4.%284%29+-12%7D+%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%0A%5Cdisplaystyle+%5Clim_%7B%28x%2Cy%29+%5Cto+%282%2C4%29%7D++%5Cfrac%7B4%2B4%7D%7B+16-12%2B16+-12%7D+%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%0A%0A%0A%5Cdisplaystyle+%5Clim_%7B%28x%2Cy%29+%5Cto+%282%2C4%29%7D++%5Cfrac%7B8%7D%7B+32-24%7D+%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%0A%0A%5Cdisplaystyle+%5Clim_%7B%28x%2Cy%29+%5Cto+%282%2C4%29%7D++%5Cfrac%7B8%7D%7B8%7D+%0A%0A%0A%0A%0A%0A "\displaystyle \lim_{(x,y) \to (2,4)} \frac{y+4}{ x^{2} y-3y+4 x^{2} -12} \\\\\\\\\\\
\displaystyle \lim_{(x,y) \to (2,4)} \frac{4+4}{ (2)^{2} .4-3(4)+4.(2)^{2} -12} \\\\\\\\\\\
\displaystyle \lim_{(x,y) \to (2,4)} \frac{4+4}{ 4.4-3(4)+4.(4) -12} \\\\\\\\\\\
\displaystyle \lim_{(x,y) \to (2,4)} \frac{4+4}{ 16-12+16 -12} \\\\\\\\\\\
\displaystyle \lim_{(x,y) \to (2,4)} \frac{8}{ 32-24} \\\\\\\\\\\
\displaystyle \lim_{(x,y) \to (2,4)} \frac{8}{8}")
Boa tarde!
Bons estudos!
Para resolver esse limite,basta substituir os valores,pois o mesmo não apresenta indeterminação é só fazer a substituição.
Boa tarde!
Bons estudos!
Usuário anônimo:
Qualquer dúvida comente!
Sem dúvidas ! Muito obrigado amigo ^^ !
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