I have the following limit:$$L=\lim_{(x,y) \to (1,1)} \dfrac{x^2+y^2-2}{|x-1|+|y-1|}$$ If I take the limit for $x=1$, then we get:$$L_1=\lim_{y \to 1_{+}}\dfrac{y^2-1}{|y-1|}=2$$, and $$L_2=\lim_{y \to 1_{-}}\dfrac{y^2-1}{|y-1|}=-2$$ So, because $L_1 \neq L_2$ does this mean that $L$ does not exist? If not how to solve the limit. I cannot use polar coordinates, nor the path $x=y$ (I get a similar result from the one cited above)
For the x=a path, the limit doesn't exist. Does the multivariable limit exist?
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calculus
limits
multivariable-calculus
absolute-value
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1if multiple values of a limit can be achieved, it has no limit. – 2017-01-09
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2Yes, that means the two-variable limit does not exist. – 2017-01-09
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0amazing. Also if on a path, one limit gives me infinity and the other a certain value is the same, right? Only the case when they are finite and equal the multivariable exists – 2017-01-09
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0@DenisNichita: Yes and no. If the limit exists, it will be the same along all paths. So if you find two paths along which the limits disagree (or one does not exist), the two-variable limit does not exist. But choosing two paths and showing the limits along those paths agree is not sufficient to show the limit exists. – 2017-01-09
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0Things like squeeze theorem are usually used to secure these limits in general. – 2017-01-23