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AP Calculus AB : Understanding continuity in terms of limits

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Example Questions

Example Question #2 : Continunity As A Property Of Functions

If $\lim_{x \to 0} f(x)$ exists,

Possible Answers:

f(0) exists and $\lim_{x \to 0}f(x) = f(0)$

f(x) must be continuous at all values.

We cannot conclude any of the other answers.

f(x) must be continuous at x=0 .

$\lim_{x \to \infty} f(x)$ exists.

Correct answer:

We cannot conclude any of the other answers.

Explanation:

Unless we are explicitly told so, via graph, information, or otherwise, we cannot assume f(x) is continuous at x=0 unless $\lim_{x \to 0}f(x) = f(0)$ , which is required for f(x) to be continuous at x=0 .

We cannot assume anything about the existence of $\lim_{x \to \infty} f(x)$ , because we do not know what f(x) is, or its end behavior.

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Example Question #4 : Continunity As A Property Of Functions

$f(x) = \left\{\begin{matrix} x^{2}- 3x+ 17,& x<2 \\ 17, & x=2\\ x^{2}- 4x+ 16,&x > 2 \end{matrix}\right.$

Which of the following is equal to $\lim_{x\rightarrow 2 }f(x)$ ?

Possible Answers:

$\lim_{x\rightarrow 2 }f(x)$ does not exist.

13.5

Correct answer:

$\lim_{x\rightarrow 2 }f(x)$ does not exist.

Explanation:

The limit of a function as approaches a value $x_{0}$ exists if and only if the limit from the left is equal to the limit from the right; the actual value of $f(x_{0})$ is irrelevant. Since the function is piecewise-defined, we can determine whether these limits are equal by finding the limits of the individual expressions. These are both polynomials, so the limits can be calculated using straightforward substitution:

$\lim_{x\rightarrow 2^{-} }f(x)= \lim_{x\rightarrow 2^{-} } (x^{2}- 3x+ 17) = 2^{2}- 3(2)+ 17 = 15$

$\lim_{x\rightarrow 2^{+} }f(x) = \lim_{x\rightarrow 2^{+} }(x^{2}- 4x+ 16) = 2^{2}- 4(2)+16 = 12$

$\lim_{x\rightarrow 2 }f(x)$ does not exist, because $\lim_{x\rightarrow 2^{-} }f(x) \ne \lim_{x\rightarrow 2^{+} }f(x)$ .

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Example Question #2 : Continunity As A Property Of Functions

Determine any points of discontinuity for the function:

$f(x)=\frac{2}{ln(x^2)}$

Possible Answers:

x=-1,1

x=0

x=-1,0.1

x=0, 1

x=-1

Correct answer:

x=-1,1

Explanation:

For a function to be continuous the following criteria must be met:

The function must exist at the point (no division by zero, asymptotic behavior, negative logs, or negative radicals).
The limit must exist.
The point must equal the limit. (Symbolically, $\lim_{x\rightarrow c}f(x)=f(c)$ ).

It is easiest to first find any points where the function is undefined. Since our function involves a fraction and a natural log, we must find all points in the domain such that the natural log is less than or equal to zero, or points where the denominator is equal to zero.

To find the values that cause the natural log to be negative we set

ln(x^2)=0

$e^{ln(x^2)}=e^{0}$

x^2=1

$x=\pm1$

Therefore, those x values will yield our points of discontinuity. Normally, we would find values where the natural log is negative; however, for all x^2 the function is positive.

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AP Calculus AB : Understanding continuity in terms of limits

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #2 : Continunity As A Property Of Functions

Example Question #4 : Continunity As A Property Of Functions

Example Question #2 : Continunity As A Property Of Functions

All AP Calculus AB Resources

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