What a limit of two variables means
For one variable, \(x\) can approach \(a\) from only two directions: the left and the right. For two variables, the point \((x, y)\) can approach \((a, b)\) along infinitely many paths: straight lines, parabolas, spirals. That is what makes these limits harder.
\(\displaystyle\lim_{(x, y) \to (a, b)} f(x, y) = L\) if \(f(x, y)\) gets as close to \(L\) as you like whenever \((x, y)\) is close enough to \((a, b)\), whatever the path. Formally: for every \(\varepsilon > 0\) there is a \(\delta > 0\) with \[|f(x, y) - L| < \varepsilon \quad \text{whenever} \quad 0 < \sqrt{(x - a)^2 + (y - b)^2} < \delta\]
This limit is also called the double or simultaneous limit.
The idea in one picture
Three tools
1. Show a limit does not exist: find two paths that disagree
Try straight lines \(y = mx\). If the answer depends on \(m\), the limit doesn't exist. If every line gives the same answer, that proves nothing yet: also try a curve such as \(y = mx^2\) or \(x = my^2\).
2. Show a limit exists: switch to polar coordinates
Put \(x = r\cos\theta\), \(y = r\sin\theta\). Then \((x, y) \to (0, 0)\) means \(r \to 0\), whatever \(\theta\) does. If you can show \(|f - L| \le g(r)\), where \(g(r) \to 0\) does not depend on \(\theta\), the limit is \(L\).
3. Repeated limits
The repeated (iterated) limits take one variable at a time: \(\displaystyle\lim_{x \to a}\Big[\lim_{y \to b} f(x, y)\Big]\) and \(\displaystyle\lim_{y \to b}\Big[\lim_{x \to a} f(x, y)\Big]\).
- If the two repeated limits are different, the double limit does not exist.
- But if they are equal, the double limit still may not exist (Example 3).
Continuity
\(f(x, y)\) is continuous at \((a, b)\) if
1.
\(f(a, b)\) is defined,
2.
\(\displaystyle\lim_{(x, y) \to (a, b)} f(x, y)\) exists, and
3.
the limit equals \(f(a, b)\).
Polynomials, \(e^x\), \(\sin\), \(\cos\), and their sums, products and compositions are continuous wherever they are defined. Exam questions therefore focus on the one awkward point, usually the origin, of a function defined in two pieces.
A surprise: unlike one-variable calculus, a function can have both partial derivatives at a point and still be discontinuous there (Example 6). Having partial derivatives only says the function behaves well along the two axis directions.
Solved examples
Example 1: Direct substitution
EasyEvaluate \(\displaystyle\lim_{(x, y) \to (1, 2)}\frac{x^2 + xy}{x + y}\).
- The function is a quotient of polynomials, and the denominator \(1 + 2 \ne 0\) at the point.
- So substitute: \(\dfrac{1 + 2}{3}\).
Answer\(1\)
Example 2: Proving a limit exists (polar)
MediumShow that \(\displaystyle\lim_{(x, y) \to (0, 0)}\frac{x^2 y}{x^2 + y^2} = 0\).
- Put \(x = r\cos\theta\), \(y = r\sin\theta\): \(\dfrac{r^3\cos^2\theta\sin\theta}{r^2} = r\cos^2\theta\sin\theta\).
- \(|r\cos^2\theta\sin\theta| \le r\), and \(r \to 0\) regardless of \(\theta\).
AnswerThe limit exists and equals \(0\).
Example 3: A limit that does not exist
MediumDoes \(\displaystyle\lim_{(x, y) \to (0, 0)}\frac{xy}{x^2 + y^2}\) exist?
- Along \(y = mx\): \(\dfrac{mx^2}{x^2 + m^2x^2} = \dfrac{m}{1 + m^2}\).
- This depends on \(m\): \(0\) along \(y = 0\), but \(\tfrac{1}{2}\) along \(y = x\).
- Notice that both repeated limits are 0, yet the double limit doesn't exist.
AnswerNo.
Example 4: The parabola trap
Exam levelDoes \(\displaystyle\lim_{(x, y) \to (0, 0)}\frac{x^2 y}{x^4 + y^2}\) exist?
- Along \(y = mx\): \(\dfrac{mx^3}{x^4 + m^2x^2} = \dfrac{mx}{x^2 + m^2} \to 0\) for every \(m\). Along \(x = 0\) it is also 0.
- So every straight line gives 0. But try the parabola \(y = x^2\): \(\dfrac{x^4}{x^4 + x^4} = \dfrac{1}{2}\).
- Two paths, two different answers (see the diagram above).
AnswerNo. Equal limits along every straight line are not enough.
Example 5: Unequal repeated limits
MediumFind the repeated limits of \(f(x, y) = \dfrac{x - y}{x + y}\) at \((0, 0)\).
- \(\displaystyle\lim_{x \to 0}\Big[\lim_{y \to 0}\frac{x - y}{x + y}\Big] = \lim_{x \to 0}\frac{x}{x} = 1\).
- \(\displaystyle\lim_{y \to 0}\Big[\lim_{x \to 0}\frac{x - y}{x + y}\Big] = \lim_{y \to 0}\frac{-y}{y} = -1\).
AnswerThe repeated limits are \(1\) and \(-1\), so the double limit does not exist.
Example 6: Partial derivatives exist, but not continuous
Exam levelLet \(f(x, y) = \dfrac{xy}{x^2 + y^2}\) for \((x, y) \ne (0, 0)\), and \(f(0, 0) = 0\). Is \(f\) continuous at the origin? Do \(f_x(0, 0)\) and \(f_y(0, 0)\) exist?
- From Example 3, the limit at the origin doesn't exist, so \(f\) is not continuous there.
- \(f_x(0, 0) = \displaystyle\lim_{h \to 0}\frac{f(h, 0) - f(0, 0)}{h} = \lim_{h \to 0}\frac{0 - 0}{h} = 0\). In the same way, \(f_y(0, 0) = 0\).
AnswerNot continuous, yet both partial derivatives exist (and equal 0).
Example 7: Checking continuity
MediumLet \(f(x, y) = \dfrac{x^2 y^2}{x^2 + y^2}\) for \((x, y) \ne (0, 0)\), and \(f(0, 0) = 0\). Is \(f\) continuous at the origin?
- Polar: \(\dfrac{r^4\cos^2\theta\sin^2\theta}{r^2} = r^2\cos^2\theta\sin^2\theta\), and \(|r^2\cos^2\theta\sin^2\theta| \le r^2 \to 0\).
- So the limit is \(0\), which equals \(f(0, 0)\).
AnswerYes, \(f\) is continuous at the origin.
Decide quickly which way to go. If the top and bottom have the same total degree (like \(\dfrac{xy}{x^2 + y^2}\), degree 2 over 2), suspect the limit doesn't exist and try \(y = mx\). If the top has the higher degree (like \(\dfrac{x^2 y}{x^2 + y^2}\), 3 over 2), suspect it does and use polar coordinates. Mixed powers like \(x^4\) and \(y^2\) are a hint to try a parabola.
Common mistakes
1. Trying only straight lines and declaring the limit exists
Lines can all agree while a curve disagrees (Example 4). Agreement along paths can only disprove a limit, never prove one. To prove it, use polar coordinates or a bound.
2. A polar bound that still depends on \(\theta\)
If after substituting you're left with something like \(\dfrac{\cos\theta\sin\theta}{\cos^2\theta}\) with no factor of \(r\), the expression doesn't go to a single value, so the limit doesn't exist.
3. Confusing repeated limits with the double limit
Equal repeated limits do not prove that the double limit exists.
4. Thinking partial derivatives imply continuity
True for one variable (differentiable implies continuous), but false for partial derivatives (Example 6).
How it's asked in exams
- Evaluate a limit, or show it does not exist.
- Find the repeated limits and comment on the double limit.
- Test continuity of a two-piece function at the origin.
- Show that the partial derivatives exist but the function is not continuous (Example 6).
Practice questions
Try each one on paper before you open the answer.
Q1\(\displaystyle\lim_{(x, y) \to (0, 0)}\frac{x^3 + y^3}{x^2 + y^2}\)
Polar: \(r(\cos^3\theta + \sin^3\theta)\), and \(|\dots| \le 2r \to 0\). The limit is \(0\).
Q2\(\displaystyle\lim_{(x, y) \to (0, 0)}\frac{x^2 - y^2}{x^2 + y^2}\)
Along \(y = mx\): \(\dfrac{1 - m^2}{1 + m^2}\), which depends on \(m\). The limit does not exist.
Q3\(\displaystyle\lim_{(x, y) \to (0, 0)}\frac{\sin(x^2 + y^2)}{x^2 + y^2}\)
Let \(t = x^2 + y^2 \to 0\). Then \(\dfrac{\sin t}{t} \to 1\). The limit is \(1\).
Q4\(\displaystyle\lim_{(x, y) \to (0, 0)}\frac{xy^2}{x^2 + y^4}\)
Lines \(y = mx\) give \(\dfrac{m^2x}{1 + m^4x^2} \to 0\). But along \(x = y^2\): \(\dfrac{y^4}{2y^4} = \tfrac{1}{2}\). The limit does not exist.
Q5Is \(f(x, y) = \dfrac{2x^2 y}{x^2 + y^2}\), with \(f(0, 0) = 0\), continuous at the origin?
Polar: \(2r\cos^2\theta\sin\theta\), and \(|\dots| \le 2r \to 0 = f(0, 0)\). Yes, it is continuous.
Q6Find the repeated limits of \(\dfrac{x^2 - y^2}{x^2 + y^2}\) at the origin.
\(\displaystyle\lim_{x \to 0}\lim_{y \to 0} = \lim_{x \to 0}\frac{x^2}{x^2} = 1\) and \(\displaystyle\lim_{y \to 0}\lim_{x \to 0} = \lim_{y \to 0}\frac{-y^2}{y^2} = -1\). They differ, so the double limit does not exist.
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Frequently asked questions
Why are two-variable limits harder than one-variable limits?
Because there are infinitely many paths to approach a point, and the limit must be the same along all of them.
How do I prove a two-variable limit does not exist?
Find two paths that give different values. Lines \(y = mx\) are the first thing to try; if they all agree, try a curve such as \(y = mx^2\).
How do I prove a two-variable limit exists?
Usually by switching to polar coordinates and bounding the expression by something that depends only on \(r\) and tends to 0.
If a function has partial derivatives at a point, is it continuous there?
Not necessarily. \(\dfrac{xy}{x^2 + y^2}\) (with value 0 at the origin) has both partial derivatives at the origin but is not continuous there.
Where this leads next
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Vipul Sir has taught mathematics for over 15 years to HSC, CBSE, ICSE, IGCSE/IB, Diploma, Engineering and BSc students at VVS Classes in Goregaon and Vile Parle, Mumbai. He teaches every class himself, concepts first and exam technique second.