Calculating limits using algebra - AP Calculus AB

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Question

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

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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

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

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

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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:

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

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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:

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

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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:

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

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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:

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

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

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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

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

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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

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

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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

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

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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

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

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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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Question

$\begin{align}&\text{Evaluate the limit of }\frac{4x + x^{2} - 60}{2x + x^{2} - 80}\&\text{At the point }x=-10\end{align}$

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Answer

$\begin{align}&\text{Simply plugging in the value of }x=-10\text{ into the function}\&\frac{4x + x^{2} - 60}{2x + x^{2} - 80}\&\text{will not tell us much, because it simply gives us a zero in}\&\text{the numerator and denominator: }\frac{0}{0}\&\text{Note, however, that we're calculating a limit. As such, we can try to simplify our fraction.}\&\text{We can do this because we're not actually evaluating the function at these zero points,}\&\text{just close to them. Factoring our function we find:}\&\frac{(x - 6)\cdot (x + 10)}{(x - 8)\cdot (x + 10)}\&\text{Which reduces to:}\&\frac{(x - 6)}{(x - 8)}\&\text{At the value of }x=-10\text{ the function is}\&\frac{8}{9}\end{align}$

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Question

$\begin{align}&\text{Find the limit of }f(x)=\frac{42x + 3x^{2} + 144}{43x + 5x^{2} + 24}\&\text{At the location }x=-8\end{align}$

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Answer

$\begin{align}&\text{Simply plugging in the value of }x=-8\text{ into the function}\&\frac{42x + 3x^{2} + 144}{43x + 5x^{2} + 24}\&\text{will not tell us much, because it simply gives us a zero in}\&\text{the numerator and denominator: }\frac{0}{0}\&\text{Note, however, that we're calculating a limit. As such, we can try to simplify our fraction.}\&\text{We can do this because we're not actually evaluating the function at these zero points,}\&\text{just close to them. Factoring our function we find:}\&\frac{3\cdot (x + 6)\cdot (x + 8)}{(5x + 3)\cdot (x + 8)}\&\text{Which reduces to:}\&\frac{(3x + 18)}{(5x + 3)}\&\text{At the value of }x=-8\text{ the function is}\&\frac{6}{37}\end{align}$

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Question

$\begin{align}&\text{Find the limit of }f(x)=\frac{x^{2} - 32x + 252}{x^{2} - 20x + 84}\&\text{At the location }x=14\end{align}$

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Answer

$\begin{align}&\text{Simply plugging in the value of }x=14\text{ into the function}\&\frac{x^{2} - 32x + 252}{x^{2} - 20x + 84}\&\text{will not tell us much, because it simply gives us a zero in}\&\text{the numerator and denominator: }\frac{0}{0}\&\text{Note, however, that we're calculating a limit. As such, we can try to simplify our fraction.}\&\text{We can do this because we're not actually evaluating the function at these zero points,}\&\text{just close to them. Factoring our function we find:}\&\frac{(x - 14)\cdot (x - 18)}{(x - 6)\cdot (x - 14)}\&\text{Which reduces to:}\&\frac{(x - 18)}{(x - 6)}\&\text{At the value of }x=14\text{ the function is}\&-\frac{1}{2}\&\text{Note: To find roots, you can always utilize the quadratic formula:}\&ax^2+bx+c\&x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\end{align}$

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Question

$\begin{align}&\text{Determine the limit if it exists of }\frac{9x + 2x^{2} - 35}{59x + 5x^{2} + 168}\&\text{At }x=-7\end{align}$

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Answer

$\begin{align}&\text{Simply plugging in the value of }x=-7\text{ into the function}\&\frac{9x + 2x^{2} - 35}{59x + 5x^{2} + 168}\&\text{will not tell us much, because it simply gives us a zero in}\&\text{the numerator and denominator: }\frac{0}{0}\&\text{Note, however, that we're calculating a limit. As such, we can try to simplify our fraction.}\&\text{We can do this because we're not actually evaluating the function at these zero points,}\&\text{just close to them. Factoring our function we find:}\&\frac{(2x - 5)\cdot (x + 7)}{(5x + 24)\cdot (x + 7)}\&\text{Which reduces to:}\&\frac{(2x - 5)}{(5x + 24)}\&\text{At the value of }x=-7\text{ the function is}\&\frac{19}{11}\&\text{Note: To find roots, you can always utilize the quadratic formula:}\&\text{For equations of the form: }ax^2+bx+c\&\text{The roots are: }x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\end{align}$

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Question

$\begin{align}&\text{Find the limit of }f(x)=\frac{x + 23}{112x + 5x^{2} - 69}\&\text{At the location }x=-23\end{align}$

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Answer

$\begin{align}&\text{Simply plugging in the value of }x=-23\text{ into the function}\&\frac{x + 23}{112x + 5x^{2} - 69}\&\text{will not tell us much, because it simply gives us a zero in}\&\text{the numerator and denominator: }\frac{0}{0}\&\text{Note, however, that we're calculating a limit. As such, we can try to simplify our fraction.}\&\text{We can do this because we're not actually evaluating the function at these zero points,}\&\text{just close to them. Factoring our function we find:}\&\frac{x + 23}{(5x - 3)\cdot (x + 23)}\&\text{Which reduces to:}\&\frac{1}{(5x - 3)}\&\text{At the value of }x=-23\text{ the function is}\&-\frac{1}{118}\&\text{Note: To find roots, you can always utilize the quadratic formula:}\&\text{For equations of the form: }ax^2+bx+c\&\text{The roots are: }x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\end{align}$

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Question

$\begin{align}&\text{Find the limit of }f(x)=\frac{5x^{2} - 41x - 36}{x^{2} - 21x + 108}\&\text{At the location }x=9\end{align}$

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Answer

$\begin{align}&\text{Simply plugging in the value of }x=9\text{ into the function}\&\frac{5x^{2} - 41x - 36}{x^{2} - 21x + 108}\&\text{will not tell us much, because it simply gives us a zero in}\&\text{the numerator and denominator: }\frac{0}{0}\&\text{Note, however, that we're calculating a limit. As such, we can try to simplify our fraction.}\&\text{We can do this because we're not actually evaluating the function at these zero points,}\&\text{just close to them. Factoring our function we find:}\&\frac{(5x + 4)\cdot (x - 9)}{(x - 9)\cdot (x - 12)}\&\text{Which reduces to:}\&\frac{(5x + 4)}{(x - 12)}\&\text{At the value of }x=9\text{ the function is}\&-\frac{49}{3}\&\text{Note: To find roots, you can always utilize the quadratic formula:}\&\text{For equations of the form: }ax^2+bx+c\&\text{The roots are: }x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\end{align}$

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Question

$\begin{align}&\text{Determine the limit if it exists of }\frac{73x + 3x^{2} + 342}{52x + 4x^{2} - 360}\&\text{At }x=-18\end{align}$

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Answer

$\begin{align}&\text{Simply plugging in the value of }x=-18\text{ into the function}\&\frac{73x + 3x^{2} + 342}{52x + 4x^{2} - 360}\&\text{will not tell us much, because it simply gives us a zero in}\&\text{the numerator and denominator: }\frac{0}{0}\&\text{Note, however, that we're calculating a limit. As such, we can try to simplify our fraction.}\&\text{We can do this because we're not actually evaluating the function at these zero points,}\&\text{just close to them. Factoring our function we find:}\&\frac{(3x + 19)\cdot (x + 18)}{4\cdot (x - 5)\cdot (x + 18)}\&\text{Which reduces to:}\&\frac{(3x + 19)}{(4x - 20)}\&\text{At the value of }x=-18\text{ the function is}\&\frac{35}{92}\&\text{Note: To find roots, you can always utilize the quadratic formula:}\&\text{For equations of the form: }ax^2+bx+c\&\text{The roots are: }x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\end{align}$

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Question

$\begin{align}&\text{Evaluate the limit of }\frac{3x^{2} - 24x + 36}{x - 6}\&\text{At the point }x=6\end{align}$

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Answer

$\begin{align}&\text{Simply plugging in the value of }x=6\text{ into the function}\&\frac{3x^{2} - 24x + 36}{x - 6}\&\text{will not tell us much, because it simply gives us a zero in}\&\text{the numerator and denominator: }\frac{0}{0}\&\text{Note, however, that we're calculating a limit. As such, we can try to simplify our fraction.}\&\text{We can do this because we're not actually evaluating the function at these zero points,}\&\text{just close to them. Factoring our function we find:}\&\frac{3\cdot (x - 2)\cdot (x - 6)}{x - 6}\&\text{Which reduces to:}\&3x - 6\&\text{At the value of }x=6\text{ the function is}\&12\&\text{Note: To find roots, you can always utilize the quadratic formula:}\&\text{For equations of the form: }ax^2+bx+c\&\text{The roots are: }x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\end{align}$

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