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How does the slope parameter k determine the limit of f(x,y)f(x,y)=xy/(x²+y²) along the line y=kx?

Along the line y=kxy=kx, substituting into the function yields f(x,kx)=k1+k2f(x,kx) = \frac{k}{1+k^2} for x≠0x \neq 0. This value depends on the slope kk and is constant with respect to xx. Because the limit varies with kk (e.g., k=0k=0 gives 0, k=1k=1 gives 1/21/2), the multivariable limit does not exist.

Conditions

  • The function is f(x,y)=xy/(x2+y2)f(x,y)=xy/(x^2+y^2).
  • The path is the straight line y=kxy=kx.
  • x≠0x \neq 0.

Reasoning, step by step

  1. Substitute y=kxy=kx into the function: f(x,kx)=x(kx)x2+(kx)2f(x,kx) = \frac{x(kx)}{x^2+(kx)^2}.
  2. Simplify the expression: kx2x2(1+k2)\frac{kx^2}{x^2(1+k^2)}.
  3. Cancel x2x^2 (valid since x≠0x \neq 0): k1+k2\frac{k}{1+k^2}.
  4. Observe that the result is independent of xx but depends on kk.
  5. Conclude that since different values of kk yield different limits, the overall limit does not exist.

Example

The card states: 'Along y=kx the value is k/(1+k1+k²), which varies with k and is zero for k=0k=0.' Formula: f(x,kx)=k1+k2(x≠0)f(x,kx)=\frac{k}{1+k^2}\qquad(x\ne0).

Common misconceptions

  • Believing that because the limit along any single straight line exists, the multivariable limit exists; the video clarifies that agreement along all straight lines alone need not prove one.
  • Forgetting the condition x≠0x \neq 0 when simplifying the expression.
  • Assuming the limit is always zero because the numerator has an xx term.

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Answers are generated from source material and independently checked. Consult the original video or creator if something is unclear.