Question

Nonexistence of a limit Investiage the limit lim_((x,y)->(0,0)) (x+y)^2/(x^2+y^2)

Multivariable functions
ANSWERED
asked 2021-01-31
Nonexistence of a limit Investiage the limit \(\displaystyle\lim_{{{\left({x},{y}\right)}\to{\left({0},{0}\right)}}}\frac{{\left({x}+{y}\right)}^{{2}}}{{{x}^{{2}}+{y}^{{2}}}}\)

Answers (1)

2021-02-01
By applying identity \(\displaystyle{\left({a}+{b}\right)}^{{2}}={\left({a}^{{2}}+{b}^{{2}}+{2}{a}{b}\right)}\)
\(\displaystyle\lim_{{{\left({x},{y}\right)}\to{\left({0},{0}\right)}}}\frac{{\left({x}+{y}\right)}^{{2}}}{{{x}^{{2}}+{y}^{{2}}}}=\lim_{{{\left({x},{y}\right)}\to{\left({0},{0}\right)}}}\frac{{{x}^{{2}}+{y}^{{2}}+{2}{x}{y}}}{{{x}^{{2}}+{y}^{{2}}}}\)
\(\displaystyle\lim_{{{\left({x},{y}\right)}\to{\left({0},{0}\right)}}}\frac{{\left({x}+{y}\right)}^{{2}}}{{{x}^{{2}}+{y}^{{2}}}}=\lim_{{{\left({x},{y}\right)}\to{\left({0},{0}\right)}}}\frac{{{x}^{{2}}+{y}^{{2}}}}{{{x}^{{2}}+{y}^{{2}}}}+\frac{{{2}{x}{y}}}{{{x}^{{2}}+{y}^{{2}}}}\)
\(\displaystyle\lim_{{{\left({x},{y}\right)}\to{\left({0},{0}\right)}}}\frac{{\left({x}+{y}\right)}^{{2}}}{{{x}^{{2}}+{y}^{{2}}}}=\lim_{{{\left({x},{y}\right)}\to{\left({0},{0}\right)}}}\frac{{{x}^{{2}}+{y}^{{2}}}}{{{x}^{{2}}+{y}^{{2}}}}+\lim_{{{\left({x},{y}\right)}\to{\left({0},{0}\right)}}}\frac{{{2}{x}{y}}}{{{x}^{{2}}+{y}^{{2}}}}\)
\(\displaystyle\lim_{{{\left({x},{y}\right)}\to{\left({0},{0}\right)}}}\frac{{\left({x}+{y}\right)}^{{2}}}{{{x}^{{2}}+{y}^{{2}}}}=\lim_{{{\left({x},{y}\right)}\to{\left({0},{0}\right)}}}{1}+\lim_{{{\left({x},{y}\right)}\to{\left({0},{0}\right)}}}\frac{{{2}{x}{y}}}{{{x}^{{2}}+{y}^{{2}}}}\)
\(\displaystyle\lim_{{{\left({x},{y}\right)}\to{\left({0},{0}\right)}}}\frac{{\left({x}+{y}\right)}^{{2}}}{{{x}^{{2}}+{y}^{{2}}}}={1}+{0}\)
\(\displaystyle\lim_{{{\left({x},{y}\right)}\to{\left({0},{0}\right)}}}\frac{{\left({x}+{y}\right)}^{{2}}}{{{x}^{{2}}+{y}^{{2}}}}={1}\)
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