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Limits Around Discontinuities

April 8th, 2009 |

Author: Merrin

Suppose f(x) is an otherwise continuous function, but has a point removed. Such a discontinuity is called a removable discontinuity because if the point is returned there is a continuous function again. Here is one way to zap a point. Let g(x) be defined as

$g(x)=\left( \cfrac{x-c}{x-c} \right)f(x)$

The point at x=c cannot be in the domain of g(x) because that would imply division by zero. The factor in front can be canceled otherwise when x is not equal to c.

\includegraphics[width= 5in]{../images/FunctionHole.pdf}

\caption{A hole in an otherwise continuous function is drawn with an open circle.}

Example 1. With f(x) everywhere a continuous function and g(x) defined as above, calculate

$\mathop{\lim }\limits_{x\to c} g(x)=\mathop{\lim }\limits_{x\to c} \left(\cfrac{x-c}{x-c} \right)f(x)$

Solution 1. We can use the properties of limits and continuity of f(x) to write

$\mathop{\lim }\limits_{x\to c} g(x)=\mathop{\lim }\limits_{x\to c} \cfrac{x-c}{x-c} f(x)=\mathop{\lim }\limits_{x\to c} \cfrac{x-c}{x-c} \mathop{\lim }\limits_{x\to c} f(x)=\mathop{\lim }\limits_{x\to c} \cfrac{x-c}{x-c} \mathop{\lim }\limits_{x\to c} f(c)=f(c)$

When in doubt, calculate the left and right hand limits and see if they are equal.

${\mathop{\lim }\limits_{x\to c^{+} } \cfrac{x-c}{x-c} =\mathop{\lim }\limits_{\varepsilon \to 0^{+} } \cfrac{c+\varepsilon -c}{c+\varepsilon -c} =\mathop{\lim }\limits_{\varepsilon \to 0^{+} } \cfrac{\varepsilon }{\varepsilon } =1} \\ {\mathop{\lim }\limits_{x\to c^{-} } \cfrac{x-c}{x-c} =\mathop{\lim }\limits_{\varepsilon \to 0^{+} } \cfrac{c-\varepsilon -c}{c-\varepsilon -c} =\mathop{\lim }\limits_{\varepsilon \to 0^{+} } \cfrac{-\varepsilon }{-\varepsilon } =1}$

Again I have I have used in the first equation the fact that

$\mathop{\lim }\limits_{x\to c^{+} } f(x)=\mathop{\lim }\limits_{\varepsilon \to 0^{+} } f(c+\varepsilon )$

This substitution with epsilon always as a positive number can help us keep the sign of the calculation straight. Since the left hand and right hand limits are equal and have the value one we can say.

$\mathop{\lim }\limits_{x\to c} g(x)=f(c)$

We have found a situation where there is a hole in a function but we get the same result as if the function was continuous there. Also since we have shown that the limit of a product of functions is the product of limits, then we have proved a useful theorem which we can call the cancellation theorem.

Theorem 1. If f(x) is any function such that

$\mathop{\lim }\limits_{x\to c} f(x)$

exists then

$\mathop{\lim }\limits_{x\to c} \cfrac{x-c}{x-c} f(x)=\mathop{\lim}\limits_{x\to c} f(x)$

Example 2. If g(x) is the function

$g(x)=\mathop{\lim }\limits_{x\to c} \cfrac{(x-c)^{3} }{(x-c)^{3} } f(x)$

Solution 2. Since the limit of a product can be written as a product of limits we can write.

${\mathop{\lim }\limits_{x\to c} \cfrac{(x-c)^{3} }{(x-c)^{3} } f(x)} = {\mathop{\lim }\limits_{x\to c} \cfrac{x-c}{x-c} \mathop{\lim }\limits_{x\to c} \cfrac{x-c}{x-c} \mathop{\lim }\limits_{x\to c} \cfrac{x-c}{x-c} \mathop{\lim }\limits_{x\to c} f(x)} \\ = {\mathop{\lim }\limits_{x\to c} f(x)}$

The same holds for any integer of the prefactor provided

$\mathop{\lim }\limits_{x\to c} f(x)$

exists.

Example 3. Calculate the following limit using the limit cancellation theorem

$\mathop{\lim }\limits_{x\to 1} f(x)=\mathop{\lim }\limits_{x\to 1} \cfrac{x^{3} -1}{x-1}$

Solution 3. A direct substitution of x = 1 results in an indeterminate form 0/0. This will require further analysis. First factor the expression then use the limit cancellation theorem.

${L}={\mathop{\lim }\limits_{x\to 1} \cfrac{(x-1)(x^{2} +x+1)}{(x-1)} } \\ {} ={\mathop{\lim }\limits_{x\to 1} (x^{2} +x+1)} \\={3}$

These examples should give an intuitive sense of when factors are cancelable.

Another case where we can’t use continuity to evaluate a limit is when a jump discontinuity occurs. An example of such a function with a jump discontinuity is the step function defined as

$\theta (x)=\left\{ \begin{array}{l} 1 \quad x\ge 0 \\ 0 \quad x<0 \end{array}\right.$

Example 4. Find

$\mathop{\lim }\limits_{x\to 0} \theta (x)$

Solution 4. The function jumps directly from zero to one at x = 0 so we don’t expect there to be a limit there. To prove this, we examine the one sided limits at x = 0.

$\mathop{\lim }\limits_{x\to 0^{-} } \theta (x)=0\quad \quad \mathop{\lim }\limits_{x\to 0^{+} } \theta (x)=1$