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Calculus / Chinese

Comparing function limits

Charles队长 · Bilibili · 1:18

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The explanation, unpacked.

Reviewed learning material · Video analysis · English
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The video contrasts left/right limits in one variable with approaches to a point in two variables. A multivariable limit requires uniform control of every nearby domain point. For xy/(x²+y²), the coordinate axes give zero while the diagonals give ±1/21/2, disproving the limit at the origin.

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Chapters

0:00Left and right limits0:21Approaching a point in the plane0:45Different paths disprove a limit

Learning script

Generated from the video's visuals and explanation; not verbatim speech.

Where the domain permits approach from both sides, a two-sided one-variable limit exists if the left and right limits both exist and agree. The animation first illustrates this consistency.

In two variables, the input is a point in the plane. The epsilon-delta definition requires every domain point sufficiently close to, but distinct from, the origin to give an output close to A. A few plotted paths illustrate the idea but cannot by themselves prove existence.

For f(x,y)f(x,y)=xy/(x²+y²), coordinate-axis paths give zero, y=xy=x gives 1/21/2, and y=−xy=-x gives −1/21/2. All approach the origin, but their outputs have different limits. Two differing paths suffice to disprove a limit. Agreement along all straight lines alone need not prove one.

Knowledge cards

01

Left and right limits

When both sides are available in the domain, equal existing one-sided limits give the two-sided limit.

lim⁡x→0−f(x)=lim⁡x→0+f(x)=A\lim_{x\to0^-}f(x)=\lim_{x\to0^+}f(x)=A
02

Neighborhood definition

Control every point in a punctured domain neighborhood, rather than checking a few paths or all straight lines.

∀ε>0, ∃δ>0:0<∥(x,y)∥<δ⇒∣f(x,y)−A∣<ε\forall\varepsilon>0,\ \exists\delta>0:\quad 0<\|(x,y)\|<\delta\Rightarrow|f(x,y)-A|<\varepsilon
03

Two paths can disprove

Different output limits along two approaches rule out a common limit. Existence requires uniform control of every approach.

f(x,0)=0,f(x,x)=12(x≠0)f(x,0)=0,\quad f(x,x)=\tfrac12\qquad(x\ne0)
04

The slope parameter

Along y=kx the value is k/(1+k1+k²), which varies with k and is zero for k=0k=0.

f(x,kx)=k1+k2(x≠0)f(x,kx)=\frac{k}{1+k^2}\qquad(x\ne0)

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  • Limits ExplanationAt 1:00
    Why this connection?

    The reviewed two-path card disproves a common limit using incompatible approach values for xy/(x2+y2)xy/(x^2+y^2) at the origin: the axes give 00 and the diagonals give 1/21/2 and −1/2-1/2. Establishing existence instead requires uniform control of all nearby domain points; agreement on a few paths, or even all straight lines, is not sufficient.

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