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If there are no holes in your function, it's continuous! Many powerful theorems in Calculus only apply to these special types of functions.

Which of the graphs show a function that is continous at \(x=a?\)

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\[f(x) = \frac{x^2}{(x-1)(x-2)}\]

At how many \(x\)-values is this function discontinuous?

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\[f(x)=\frac { x }{ x } ,g(x)=\frac { x^{ 2 } }{ x } ,h(x)=\frac { x }{ x^{ 2 } } \]

Each of these functions is undefined at \(x = 0.\) Which functions is it possible to extend by defining a functional value at \(x = 0\) in such a way that the resulting extended function is continuous at \(x = 0?\)

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True or False?

The function \(f(x) = x^3 + x - 1\) has a root between 0 and 1.

\[\] \[\] \[\]

**Hint:** \(f(0) = -1\) and \(f(1) = 1.\) Apply the Intermediate Value Theorem.

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Which of the graphs shows a function \(f(x)\) for which \(\lim_{x\to a} f(x)\) exists?

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