Functions, Graphs, and Limits - AP Calculus AB

Card 0 of 2277

Question

$f(x)={x^{2}+3}$ when and

$f(x)={7x-3}$ when $x\geq 3$

At the funciton described above is:

Answer

The answer is both.

If graphed the student will see that the two graphs are continuous at . There is no gap in the graph or no uneven transitions. If the graph is continuous then it is differentiable so it must be both.

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Question

Which of the following functions contains a removeable discontinuity?

Answer

A removeable discontinuity occurs whenever there is a hole in a graph that could be fixed (or "removed") by filling in a single point. Put another way, if there is a removeable discontinuity at , then the limit as approaches exists, but the value of does not.

For example, the function $f(x)=\frac{1+x^3}{1+x}$ contains a removeable discontinuity at . Notice that we could simplify as follows:

$f(x)=\frac{1+x^3}{1+x}=\frac{(1+x)(x^2-x+1)}{1+x}=x^{2}-x+1$ , where $x\neq -1$ .

Thus, we could say that $\lim_{x\rightarrow -1}\frac{1+x^3}{1+x}=\lim_{x\rightarrow -1}x^2-x+1=(-1)^2-(-1)+1=3$ .

As we can see, the limit of exists at , even though is undefined.

What this means is that will look just like the parabola with the equation $x^{2}-x+1$ EXCEPT when, where there will be a hole in the graph. However, if we were to just define , then we could essentially "remove" this discontinuity. Therefore, we can say that there is a removeable discontinuty at .

The functions

$f(x)=\frac{x}{1-x^{2}}$ , and

$f(x)=1-(\ln x)^{2}$

have discontinuities, but these discontinuities occur as vertical asymptotes, not holes, and thus are not considered removeable.

The functions

$f(x)=x^{2}\sin x^{2}$ and $f(x)=\frac{x+1}{1+x^{2}}$ are continuous over all the real values of ; they have no discontinuities of any kind.

The answer is

$f(x)=\frac{1+x^{3}}{1+x}$ .

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Question

Is the following piecewise function continuous for all ? If not, state where it is discontinuous.

$f(x) = \left{\begin{Bmatrix} x^2+3 & 3 < x \ 5x-3 & 3 \leq x < 5 \ 11(x-3) &x \geq 5 \ \end{matrix}\right.}$

Answer

To check if the piecewise function is continuous, all we need to do is check that the values at 3 and 5 line up.

At , this means checking that and have the same value. More formally, we are checking to see that $f(3) = \lim_{x\rightarrow3^-} f(x)$ , as to be continuous at a point, a function's left and right limits must both equal the function value at that point.

Plugging 3 into both, we see that both of them are 12 at . Thus, they meet up smoothly.

Next, for , we have and . Plugging in 5, we get 22 for both equations.

As all three equations are polynomials, we know they will be continuous everywhere else, and they meet up smoothly at the piecewise bounds, thus ensuring that the function is continuous everywhere.

Note, there are sharp turns at and , but this only means the function isn't differentiable at these points -- we're only concerned with continuity, which is if the equations meet up. Thus, the function is continuous.

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Question

is differentiable for which of the following values of ?

Answer

is not differentiable at and because the values are discontinuities. is not differentiable at because that point is a corner, indicating that the one-side limits at are different. is differentiable:the one side limits are the same and the point is continuous.

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Question

Consider the graph of above. What can we say about when ?

Answer

Note that , indicating that there is a horizontal tangent on at . More specifically, the derivative is the slope of the tangent line. If the slope of the tangent line is 0, then the tangent is horizontal.

The other two are incorrect because sharp turns only apply when we want to take the derivative of something. The derivative of a function at a sharp turn is undefined, meaning the graph of the derivative will be discontinuous at the sharp turn. (To see why, ask yourself if the slope at is positive 1 or negative 1?) On the other hand, integration is less picky than differentiation: We do not need a smooth function to take an integral.

In this case, to get from to , we took an integral, so it didn't matter that there was a sharp turn at the specified point. Thus, neither function had any discontinuities.

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Question

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

Answer

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

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

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Question

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

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

Answer

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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Question

Determine any points of discontinuity for the function:

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

Answer

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

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

$x=\pm1$

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

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Question

Calculate $\lim_{x \rightarrow 0}\frac{\sin 4 x^{2} }{x^{2}}$

Answer

You can substitute $u = 4x^{2}$ to write this as:

$\lim_{x \rightarrow 0}\frac{\sin 4 x^{2} }{x^{2}}$

$= \lim_{x \rightarrow 0}\left (4 \cdot \frac{\sin 4 x^{2} }{4x^{2}} \right )$

$=4 \lim_{x \rightarrow 0} \frac{\sin 4 x^{2} }{4x^{2}}$

$=4 \lim_{u \rightarrow 0} \frac{\sin u }{u}$

Note that as $x \rightarrow 0$ , $u = 4x^{2} \rightarrow 0$

$\lim_{u \rightarrow 0} \frac{\sin u }{u}$ , since the fraction becomes indeterminate, we need to take the derivative of both the top and bottom of the fraction.

$\lim_{u \rightarrow 0} \frac{\sin u }{u}=\frac{(\sin u)'}{u'}=\frac{\cos u}{1}=\cos u$

$4\lim_{u\rightarrow 0}\cos u=4\cdot \cos(0)=4\cdot 1$

$4 \lim_{u \rightarrow 0} \frac{\sin u }{u} = 4 \cdot 1 = 4$ , which is the correct choice.

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Question

Calculate $\lim_{x \rightarrow 1}\frac{\sin \left (2x^{2}-2 \right ) }{x^{2}-1}$ .

Answer

Substitute $u = 2x^{2} - 2$ to rewrite this limit in terms of u instead of x. Multiply the top and bottom of the fraction by 2 in order to make this substitution:

(Note that as $x \rightarrow1$ , $u \rightarrow 0$ .)

$\lim_{x \rightarrow 1}\frac{\sin \left (2x^{2}-2 \right ) }{x^{2}-1}$

$=2 \cdot \lim_{x \rightarrow 1}\frac{\sin \left (2x^{2}-2 \right ) }{2\left (x^{2}-1 \right )}$

$= 2 \cdot \lim_{x \rightarrow 1}\frac{\sin \left (2x^{2}-2 \right ) }{2x^{2}-2 \right )}$

$= 2 \cdot \lim_{u \rightarrow 0}\frac{\sin u } {u}$

$\lim_{u \rightarrow 0} \frac{\sin u }{u} = 1$ , so

$2 \cdot \lim_{u \rightarrow 0}\frac{\sin u } {u} = 2\cdot1 = 2$ , which is therefore the correct answer choice.

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Question

Evaluate the following limit:

$\lim_{x\rightarrow 4}\frac{x^{2}+2x-24}{x-4}$

Answer

Factor x-4 out of the numerator and simplify:

$\frac{(x-4)(x+6)}{(x-4)}=x+6$

Evaluate the limit for x=4:

Although there is a discontinuity at x=4, the limit at x=4 is 10 because the function approaches ten from the left and right side.

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Question

Evaluate the following limit:

$\lim_{x\rightarrow 2}\frac{x^2+3x-10}{x-2}$

Answer

Factor the numerator and simplify the expression.

$\frac{(x+5)(x-2)}{x-2}=x+5$

Evaluate the function at x=2.

There is a discontinuity at x=2, but since it the limit as x approaches 2 from the right is equal to the limit as x approaches 2 from the left, the limit exists.

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Question

Evaluate the following limit:

$\lim_{x\rightarrow 0}\frac{x^3+5x^2+8x}{x}$

Answer

Factor the numerator to evaluate the limit:

$\frac{x(x+2)(x+4)}{x}=(x+2)(x+4)$

Evaluate the limit:

$\lim_{x\rightarrow 0}(x+2)(x+4)=(2)(4)=8$

There is a discontinuity at x=0 but the limit is equal to 8 because the limit from the right is equal to the limit from the left.

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Question

Evaluate the following limit:

$\lim_{x\rightarrow \infty }\frac{x^3+4}{5x^2+2x+7}$

Answer

As x approaches infinity, the x values that are raised to the highest power dominate the function and all other values become negligible. In this limit, x3 has the highest order in the numerator and 5x2 has the highest power of all the values in the denominator. We can simplify the limit to the following:

$\lim_{x\rightarrow \infty }\frac{x^3}{5x^2}=\lim_{x\rightarrow \infty }\frac{x}{5}=\infty$

After factoring out x2 it is easy to see that the limit diverges to infinity.

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Question

Calculate $\lim_{x\rightarrow 5} \frac{x^2-25}{x-5}$

Answer

First we notice that substituting 5 in for x will give us a 0 in the denominator.

So we simplify the equation by noticing the numerator is the difference of two squares.

$\frac{x^2-25}{x-5}=\frac{(x+5)(x-5)}{x-5}=x+5$

Now we can substitute 5 in for x, and we arrive at our answer of 10.

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Question

$\lim_{x\rightarrow 0}\frac{x^2-2x+3}{x-2}$

Answer

The first step to evaluating limits is always attempt to direct substitute the value into the function. In this case, that is all you need to do.

$\lim_{x\rightarrow 0}\frac{x^2-2x+3}{x-2}=\frac{0^2-2(0)+3}{0-2}=\frac{-3}{2}.$

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Question

$\lim_{x\rightarrow 2}\frac{x^2+5x-14}{x-2}=?$

Answer

To start this problem, we need to factor the quadratic equation. This will result in a hole at , which will allow us to directly substitute the limit value into the resulting function to determine the limit.

$\frac{x^2+5x-14}{x-2}=\frac{(x-2)(x+7)}{x-2}=x+7.$

$\lim_{x\rightarrow 2}{x+7}=2+7=9.$

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Question

$\lim_{x\rightarrow -3}\frac{2x+4}{x-3}=?$

Answer

To evaluate this limit, we need to direct substitute the value in question into our function. Note, this value is in the domain of the function, so direct substitution is the only method we need here.

$\lim_{x\rightarrow -3}\frac{2x+4}{x-3}=\frac{2(-3)+4}{(-3-3)}=\frac{-2}{-6}=\frac{1}{3}.$

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Question

Evaluate $\lim_{x \to 3} \frac{x^2-9}{x-3}$

Answer

This limit can't be evaluated by a simple substitution; is not defined at , so some simplification is in order first.

$\lim_{x \to 3} \frac{x^2-9}{x-3}$ . (Start)

$=\lim_{x \to 3} \frac{(x+3)(x-3)}{x-3}$ . (Factor the numerator)

$=\lim_{x \to 3} x+3$ . (Cancel the terms)

. (Substitute in the )

.

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Question

$\begin{align}&\text{Determine the limit if it exists of }\frac{x - 8}{x^{2} - 21x + 104}\&\text{At }x=8\end{align}$

Answer

$\begin{align}&\text{Simply plugging in the value of }x=8\text{ into the function}\&\frac{x - 8}{x^{2} - 21x + 104}\&\text{will not tell us much, because it simply gives us a zero in}\&\text{the numerator and denominator: }\frac{0}{0}\&\text{Note, however, that we're calculating a limit. As such, we can try to simplify our fraction.}\&\text{We can do this because we're not actually evaluating the function at these zero points, just close to them.}\&\text{Factoring our function we find:}\&\frac{x - 8}{(x - 8)\cdot (x - 13)}\&\text{Which reduces to:}\&\frac{1}{(x - 13)}\&\text{At the value of }x=8\text{ the function is}\&-\frac{1}{5}\end{align}$

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