Question

By considering different paths of approach, show that the function has no limit as $(x,y)\to(0,0)$.\\ $f(x,y) = \frac{x}{\sqrt{x^2 + y^2}}$\\Find the limit as $(x,y)\to(0,0)$ along the path $y = x$ for $x > 0$.\ (Type an exact answer, using radicals as needed.)\\Find the limit as $(x,y)\to(0,0)$ along the path $y = x$ for $x < 0$.\ (Type an exact answer, using radicals as needed.)\\Why doesn't the limit exist?\\ A. The limit does not exist because $f(x,y)$ is not defined at the point $(0,0)$.\nB. The limit does not exist because $f(x,y)$ has the same limits along two different paths in the domain of $f$ as $(x,y)$ approaches $(0,0)$.\nC. The limit does not exist because $f(x,y)$ is not continuous.\nD. The limit does not exist because $f(x,y)$ has different limits along two different paths in the domain of $f$ as $(x,y)$ approaches $(0,0)$.

          By considering different paths of approach, show that the function has no limit as $(x,y)\to(0,0)$.\\
$f(x,y) = \frac{x}{\sqrt{x^2 + y^2}}$\\Find the limit as $(x,y)\to(0,0)$ along the path $y = x$ for $x > 0$.\
(Type an exact answer, using radicals as needed.)\\Find the limit as $(x,y)\to(0,0)$ along the path $y = x$ for $x < 0$.\
(Type an exact answer, using radicals as needed.)\\Why doesn't the limit exist?\\
A. The limit does not exist because $f(x,y)$ is not defined at the point $(0,0)$.\nB. The limit does not exist because $f(x,y)$ has the same limits along two different paths in the domain of $f$ as $(x,y)$ approaches $(0,0)$.\nC. The limit does not exist because $f(x,y)$ is not continuous.\nD. The limit does not exist because $f(x,y)$ has different limits along two different paths in the domain of $f$ as $(x,y)$ approaches $(0,0)$.

By considering different paths of approach, show that the function has no limit as (x,y)→(0,0).

f(x,y) = (x)/(√(x^2 + y^2))
Find the limit as (x,y)→(0,0) along the path y = x for x > 0.(Type an exact answer, using radicals as needed.)
Find the limit as (x,y)→(0,0) along the path y = x for x < 0.(Type an exact answer, using radicals as needed.)
Why doesn't the limit exist?

A. The limit does not exist because f(x,y) is not defined at the point (0,0).. The limit does not exist because f(x,y) has the same limits along two different paths in the domain of f as (x,y) approaches (0,0).. The limit does not exist because f(x,y) is not continuous.. The limit does not exist because f(x,y) has different limits along two different paths in the domain of f as (x,y) approaches (0,0).

Added by Krista A.

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