Why Translate the Coordinates?
A limit such as \(\lim_{(x,y)\to(1,2)} f(x,y)\) is about the behavior of a function near \((1,2)\), not necessarily about its value at that point. Many useful methods are easiest to organize when the target is the origin. A change of coordinates can move the target there: set \(u=x-1\) and \(v=y-2\). Then approaching \((1,2)\) in the original variables is exactly the same as approaching \((0,0)\) in the translated variables.
This is a translation, not a rotation or a rescaling. It keeps all distances unchanged. As a result, it does not alter whether a limit exists or what its value is. It changes only the coordinates used to describe the approach.
For a three-variable limit at \((a,b,c)\), use \(u=x-a\), \(v=y-b\), and \(w=z-c\). The translated function is \(g(u,v,w)=f(a+u,b+v,c+w)\). Each original coordinate must be shifted by its own target coordinate. Leaving a coordinate unshifted, or replacing it with its translated coordinate without adding back the target value, changes the function.
Why Translation Preserves a Limit
The multivariable limit definition measures how close the input is to the target. In two variables, the distance from \((x,y)\) to \((a,b)\) is \(\sqrt{(x-a)^2+(y-b)^2}\). With \(u=x-a\) and \(v=y-b\), this becomes \(\sqrt{u^2+v^2}\), the distance from \((u,v)\) to the origin. In three variables, the same calculation gives \(\sqrt{u^2+v^2+w^2}\).
Here is the full argument using the definition. Suppose the translated function \(g\) has limit \(L\) at the origin. For every \(\varepsilon>0\), there is a \(\delta>0\) such that \(0<\sqrt{u^2+v^2}<\delta\) implies \(|g(u,v)-L|<\varepsilon\). If \(0<\sqrt{(x-a)^2+(y-b)^2}<\delta\), then \(u=x-a\) and \(v=y-b\) satisfy \(0<\sqrt{u^2+v^2}<\delta\). Since \(g(u,v)=f(x,y)\), it follows that \(|f(x,y)-L|<\varepsilon\). This proves the original limit is \(L\). Reversing the substitutions proves the converse.
The proof also explains how to use the translation safely: replace every original coordinate consistently, including in denominators, exponents, and any conditions specifying where the function is defined. Once the expression is in translated coordinates, use an appropriate method for a limit at the origin.
Worked Example: Simplifying a Rational Expression
Worked Example: Simplifying a Rational Expression
Find \(\lim_{(x,y)\to(1,2)} f(x,y)\), where \(f(x,y)=\frac{(x+y-3)(x+y-1)}{x+y-3}\) for \(x+y\ne3\).
Set \(u=x-1\), \(v=y-2\). Thus \(x=1+u\) and \(y=2+v\). Substitute into each factor:
Therefore, for \(u+v\ne0\),
The cancellation is valid wherever the original function is defined; the translated domain excludes \(u+v=0\). As \((u,v)\to(0,0)\), \(2+u+v\to2\). For a direct check, given \(\varepsilon>0\), if \(\sqrt{u^2+v^2}<\delta\), then \(|u|<\delta\) and \(|v|<\delta\), so \(|g(u,v)-2|=|u+v|\le |u|+|v|<2\delta\). Choosing \(\delta=\varepsilon/2\) proves the limit is \(2\).
Returning to the original coordinates, the limit is \(2\). Substitution of \((1,2)\) into the original formula is not allowed because its denominator is zero; the calculation works by simplifying on the function's domain near the point, not by evaluating the function at the point.
Translated Coordinates and Path Tests
A path approaching \((a,b)\) becomes a path approaching the origin after translation. For example, if an original path is \(x=a+t\), \(y=b+t^2\), then the translated path is \(u=t\), \(v=t^2\). Conversely, a translated path \(u=\phi(t)\), \(v=\psi(t)\) corresponds to the original path \(x=a+\phi(t)\), \(y=b+\psi(t)\). Thus a path test at a nonzero point can be carried out either before or after translating.
If two paths give different limiting values, the multivariable limit does not exist. If several paths give the same value, that alone does not prove the limit exists; the paths test only rules out a limit when their results disagree. Translating the target does not change this logical role.
Worked Example: Finding Different Path Limits
Determine whether \(f(x,y)=\frac{x-1}{x+y-3}\), defined when \(x+y\ne3\), has a limit as \((x,y)\to(1,2)\).
Use \(u=x-1\), \(v=y-2\), so \(x=1+u\) and \(y=2+v\). The numerator becomes \(u\), and the denominator becomes
Therefore \(g(u,v)=\frac{u}{u+v}\) when \(u+v\ne0\). Along the translated path \(v=0\), \(u=t\), where \(t\ne0\),
Along the translated path \(u=0\), \(v=t\), where \(t\ne0\),
Both paths approach \((0,0)\) through points in the domain, but their function values approach different numbers. The translated limit does not exist, so the original limit at \((1,2)\) does not exist. In the original coordinates, these paths are \(y=2,\ x=1+t\), giving value \(1\), and \(x=1,\ y=2+t\), giving value \(0\). This confirms the path results after translating back.
Worked Example: Bounding a Three-Variable Limit
Worked Example: Bounding a Three-Variable Limit
Evaluate the limit of \(f(x,y,z)=\frac{(x-1)(y+2)}{1+(x-1)^2+(y+2)^2+(z-3)^2}\) as \((x,y,z)\to(1,-2,3)\).
Translate the target using \(u=x-1\), \(v=y+2\), and \(w=z-3\). Then the translated function is
The denominator is at least \(1\), so
Let \(\varepsilon>0\) and choose \(\delta=\sqrt{\varepsilon}\). If \(0<\sqrt{u^2+v^2+w^2}<\delta\), then \(|u|<\delta\) and \(|v|<\delta\). Hence
This verifies the limit definition, so the translated limit is \(0\), and therefore the original limit is \(0\). The estimate applies to every point sufficiently close to the target, not only to selected paths. Also, the original denominator is always at least \(1\), so the function is defined at the target; direct substitution gives \(0/1=0\), consistent with the proved limit.
A Common Translation Error
Every coordinate must be replaced by its target value plus its translated coordinate. Consider again \(f(x,y)=\frac{x-1}{x+y-3}\) near \((1,2)\). The correct replacements are \(x=1+u\) and \(y=2+v\). They give numerator \(u\) and denominator \(u+v\).
A specific incorrect calculation is to replace \(x\) by \(1+u\) but mistakenly replace \(y\) by \(v\), rather than by \(2+v\). That gives the erroneous expression
This expression tends to \(0\) as \((u,v)\to(0,0)\), because its denominator tends to \(-2\). But it is not the correctly translated function: it describes an incorrect substitution for \(y\). The correct full substitution gives \(\frac{u}{u+v}\), whose two path limits differ, as shown above. The error changes the behavior of the function and can produce a false conclusion.
How to Use the Translation
Translation is a preparation step, not a limit method by itself. After translating, choose a method suited to the resulting expression. Algebra may reveal a cancellation or a familiar limit. Paths may show that a limit cannot exist. A bound may show that all nearby function values must approach the same number. The translation makes the target simple; the method supplies the conclusion.
For \((a,b)\), write \(u=x-a\), \(v=y-b\). For \((a,b,c)\), also write \(w=z-c\).
Use \(x=a+u\), \(y=b+v\), and, when needed, \(z=c+w\) in the entire formula.
Note any points or curves excluded from the original function; translation moves those exclusions too.
Apply an algebraic simplification, a path test, or an appropriate bound to the translated expression.
Because translation preserves distances, the translated limit and the original limit have the same value, or both fail to exist.
The central benefit is consistency: an awkward target such as \((1,-2,3)\) becomes the origin, while the function's nearby behavior stays exactly the same. The translated variables should simplify the analysis, but they do not remove the need to check the domain, carry out substitutions correctly, or justify the chosen limit method.
Check Your Understanding
For each question, track every coordinate shift and distinguish a path test from a proof that a limit exists.
- What translated coordinates move the point \((3,-1)\) to the origin, and what are the inverse substitutions?
- If \(g(u,v)=f(2+u,-4+v)\), which point in the original coordinates corresponds to \((u,v)=(0,0)\)?
- Translate \(x+y-5\) for a limit at \((2,3)\). What expression results in \(u=x-2,\ v=y-3\)?
- Why do two different path limits for a translated function rule out the original limit?
- In three variables, what bounds on \(|u|,|v|,|w|\) follow from \(\sqrt{u^2+v^2+w^2}<\delta\)?