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15.2 Limits and Continuity

Recall if \(\lim_{x \to a} f(x) = L\) then we can make \(f(x)\) as close to \(L\) as we want by making \(x\) sufficiently close to \(a\).

If \(\lim_{x \to a} f(x) = L\) then it doesn't matter how we approach \(a\)

\[\lim_{x \to a^+} f(x) = \lim_{x \to a^-} f(x) = \lim_{x \to a} f(x) = L\]

Figure: Graph of a continuous linear function passing through point (a, f(a)) with green arrows approaching from both sides.

limit exists

Figure: Graph showing a jump discontinuity at x=a where the left and right limits do not meet.

limit DNE

If \(f(x)\) is continuous at \(x = a\), then \(\lim_{x \to a} f(x) = f(a)\)

Figure: Graph of a continuous curve with a solid dot at the point (a, f(a)).

Figure: Graph of a curve with an open circle (hole) at x=a, indicating the function is not defined there.

limit exists

but \(f(a)\) not defined

NOT continuous

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Many types of functions are continuous everywhere:

polynomial, sine, cosine, exponential

Many types are continuous wherever defined:

rational, logarithmic

Almost everything we just reviewed carry directly over to functions of two variables

\[\lim_{(x,y) \to (a,b)} f(x,y) = L\]

We can make \(f(x,y)\) as close to \(L\) as we want by making \((x,y)\) sufficiently close to \((a,b)\).

Whether \((x,y)\) can equal \((a,b)\) is not relevant for limit.

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Just like before, if \( f(x,y) \) has a limit at \( (a,b) \), it should NOT depend on how we get there.

But now we have more ways to get to \( (a,b) \).

Figure: A Cartesian coordinate system with x and y axes. Multiple colored paths (red, green, black) converge on a point (a,b).

Paths shown:

Along \( y = \frac{b}{a}x \)
Along \( y = b \)
Along \( x = a \)

There are many, many more ways! Along a parabola, exponential, sine, etc.

For limit to exist, it should NOT depend on path.

→ How can we check infinitely-many paths?

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Example

\[ \lim_{(x,y) \to (1,0)} \ln\left( \frac{1+y^2}{x^2+xy} \right) \]

We don't have to worry about path because this is a logarithmic function and it is continuous if defined.

→ If continuous, \( f(a,b) = \lim_{(x,y) \to (a,b)} f(x,y) \)

Is \( \ln\left( \frac{1+y^2}{x^2+xy} \right) \) defined at \( (1,0) \)?

Note: \( x=1, y=0 \)

\[ \ln\left( \frac{1+0}{1+0} \right) = \ln(1) = 0 \quad \text{yes!} \]

So this function is continuous at \( (1,0) \).

Therefore, \( f(1,0) = \lim_{(x,y) \to (1,0)} f(x,y) = 0 \)

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Example: Multivariable Limits

\[ \lim_{(x,y) \to (0,0)} \frac{y^2 - 4x^2}{2x^2 + y^2} \]

This is a rational function which is continuous wherever defined.

But here, it is not defined at \((0,0)\) so is not continuous at \((0,0)\).

So we can't use continuity to find limit.

Just because a function is not continuous at \((a,b)\), it does NOT automatically mean \(\lim_{(x,y) \to (a,b)} f(x,y) \text{ DNE}\).

Figure: A 2D coordinate graph showing a curve with a hole at x=a, indicating the limit exists despite the discontinuity.

Limit is still \(f(a)\)

So now we need to worry about path:

all paths should lead to same limit for limit to exist
if two paths lead to different numbers, then limit DNE

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\[ \lim_{(x,y) \to (0,0)} \frac{y^2 - 4x^2}{2x^2 + y^2} \]

Figure: Graph showing a point (a,b) approached by a vertical path x=a and a horizontal path y=b on Cartesian axes.

The two easiest paths to try are \(x = a\) and \(y = b\).

Along \(x = a = 0\)

\[ \lim_{(x,y) \to (0,0)} \frac{y^2 - 4x^2}{2x^2 + y^2} = \lim_{y \to 0} \frac{y^2}{y^2} = 1 \]

Along \(y = b = 0\)

\[ \lim_{(x,y) \to (0,0)} \frac{y^2 - 4x^2}{2x^2 + y^2} = \lim_{x \to 0} \frac{-4x^2}{2x^2} = -2 \]

Since two paths lead to different numbers, the limit at \((0,0)\) DNE.

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3D Surface Visualization: Perspective View

This figure illustrates a three-dimensional surface plot of a function \( z = f(x, y) \) over the domain \( [-1, 1] \times [-1, 1] \). The surface exhibits a complex topology with a sharp ridge or discontinuity along a central curve.

Figure: A 3D surface plot showing a complex, multi-colored landscape with a sharp central ridge on a Cartesian grid.

The color gradient on the surface represents the magnitude of the \( z \) values, transitioning from deep blue for lower values to yellow and orange for higher values. The grid lines emphasize the curvature and the steep gradients near the central feature.

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3D Surface Visualization: Frontal View

This page provides an alternative viewing angle of the same mathematical surface. This frontal perspective highlights the symmetry and the specific behavior of the function near the origin and along the primary axes.

Figure: A frontal 3D view of a mathematical surface with a distinct vertical fold and color-coded height values.

The vertical axis \( z \) ranges from approximately -2 to 1. The plot clearly shows a 'pinched' or 'folded' structure, suggesting a singularity or a point of non-differentiability where the surface converges toward the center.

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Multivariable Limits: Path-Dependent Analysis

Example

\[ \lim_{(x,y) \to (0,0)} \frac{x^2 - y^2}{x + y} \]

Clearly \( \frac{x^2 - y^2}{x + y} \) DNE (Does Not Exist) at \( (0,0) \).

Testing Specific Paths

along \( x = 0 \):
\[ \lim_{(x,y) \to (0,0)} \frac{x^2 - y^2}{x + y} = \lim_{y \to 0} \frac{-y^2}{y} = 0 \]
along \( y = 0 \):
\[ \lim_{(x,y) \to (0,0)} \frac{x^2 - y^2}{x + y} = \lim_{x \to 0} \frac{x^2}{x} = 0 \]

Figure: Coordinate axes with arrows indicating paths approaching the origin from both directions on both axes.

This is NOT enough! We need to see ALL paths lead to the same number.

Next path? \( y = x \)

along \( y = x \):

\[ \lim_{(x,y) \to (0,0)} \frac{x^2 - y^2}{x + y} = \lim_{y \to 0} \frac{y^2 - y^2}{y + y} = 0 \]

What's next? \( y = x^2 \)? \( y = x^3 \)? \( y = \sin x \)?

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We can't possibly check all paths.

So find a way to work with \( f(x,y) \) without assuming a path.

\[ \lim_{(x,y) \to (0,0)} \frac{x^2 - y^2}{x + y} = \lim_{(x,y) \to (0,0)} \frac{(x+y)(x-y)}{(x+y)} = \lim_{(x,y) \to (0,0)} (x-y) \]

goes to 0 without assuming path

goes to 0 no matter the path

= 0

without assuming a path

Limit is 0

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Multivariable Limit Example

Example

\[ \lim_{(x,y) \to (0,0)} \frac{\sin(x^2+y^2)}{x^2+y^2} \]

Clearly, continuity doesn't help.

Along \( x=0 \), \( y=0 \) same thing but not enough to conclude.

Path independent way?

Recall:

\[ \lim_{x \to 0} \frac{\sin(x)}{x} = 1 \]

Notice

As \( (x,y) \to (0,0) \), then \( x^2+y^2 \to 0 \) (no matter how we get there).

Let \( u = x^2+y^2 \)

\[ \lim_{(x,y) \to (0,0)} \frac{\sin(x^2+y^2)}{x^2+y^2} = \lim_{u \to 0} \frac{\sin(u)}{u} = 1 \]