Limits for Dummies

Shane Knight

Limits for Dummies

Shane Knight

A limit, by definition, is the Y value that is being **approached as the function gets closer to an X value. While the function can be undefined at a given X, we can still find a limit from the function by looking at what Y value is being approached to as the function reaches the X.**

The limit of a function can be determined at any given X value as long as there is not a jump, vertical asymptote, or oscillating behavior at the value.

To find the limit, one must algebraically—or graphically—evaluate the Y value when the function approaches the X.

To find the limit, one must algebraically—or graphically—evaluate the Y value when the function approaches the X.

To find a limit from a graph, you must look at the Y value that is being approached as the function "closes in" on an X value. If the function is going in two separate directions (negative & positive infinity at asymptotes) or represents a jump discontinuity, a limit cannot be found.

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Limits are not always simply variables and numeric values—trigonometric functions (sin, cosine, tangent) can also be used to derive limits.

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These three are the basic trigonometric limits based on the unit circle.

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These are the limits of sin and cosine functions taken from their respective graphs.

To remember the limits of sin and cosine functions, you must remember what their graphs look like:

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The Squeeze Theorem is used primarily to prove the existence of limit. When a Y value is not explicitly defined by the function or the graph, the squeeze theorem is used to show the limit. To utilize the squeeze theorems, two additional functions—such as a positive parabola with a reflection on the X axis—are graphed to "squeeze" the initial function. If the three limits are equivalent, the limit of the original function is proven.

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In this graph, the red line (Sin(x²)is a graphical representation of f(x), while the upper (x²) and lower (-x²) blue parabolas are g(x) and h(x) respectively. In order to prove the existence of the limit of f(x) as X approaches 0, we must utilize the squeeze theorem.

As the limit of g(x) and f(x) as X approaches 0 is equal to 0 and squeeze f(x) at 0, the limit of f(x) is proven. To put this in notation, we would write g(x) ≤ f(x) ≤ h(x)

As the limit of g(x) and f(x) as X approaches 0 is equal to 0 and squeeze f(x) at 0, the limit of f(x) is proven. To put this in notation, we would write g(x) ≤ f(x) ≤ h(x)

Limits are often found by evaluating the function without graphing it. To do this, we must solve for the limit algebraically.

For**all** limits, there are basic steps:

1) Substitution: plug the x into the function. If there is not a 0 in the denominator, the limit has been found.

2) If the limit is undefined (0 in the denominator), then the function must be factored.

3) After factoring and crossing out, substitute the x back in and solve.

For

1) Substitution: plug the x into the function. If there is not a 0 in the denominator, the limit has been found.

2) If the limit is undefined (0 in the denominator), then the function must be factored.

3) After factoring and crossing out, substitute the x back in and solve.

Here, we would simply plug in "10" to find that the limit is equal to 5.

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When using substitution for this limit, we find that there is a 0 in the denominator. However, we could factor the nominator into (x-1)(x+1) and cross out the "(x-1)" in the nominator and the "(x-1)" in the numerator, leaving x+1 in the numerator and 1 in the denominator. After plugging the x, 1, back in, we find that the limit is 2.

Not all limits can be algebraically derived from the function through substitution or factoring. There are also "special cases" that have separate methods to be used, such as trigonometric limits, limits of absolute value functions (if there is an "x" in the denominator), limits of functions with radicals in the numerator, and limits of complex fractions.

Here, we find that there is a 0 in the denominator AND that the function can not be factored. Thus, we can conclude that the limit does not exist. If graphed, the function exposes a "jump" in the line.

Limits can not be found while there is a root function in the numerator. To remove the root, we must rationalize the numerator by multiplying both the numerator and denominator by the conjugate of the numerator.

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After multiplying, we see that there is (x-4) in both the numerator and denominator, meaning we can cross out both of them while leaving a 1 in the numerator. Lastly, plug in and solve for 4 and you are left with the limit: 1/4

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To solve trigonometric functions such as the one below, we must use trigonometric substitution to re-write the function so that plugging x in does not result in "0" in the denominator.

1) Substitute "sin/cos" in place of "tan(x)," leaving a fraction in the numerator. After we move cos to the denominator, we are left with "sin(x)/xcos(x)"

2)Separate the function into two parts: sin(x)/x and 1/cos(x).

3) Evaluate the function: we know that the limit of sin(x)/x as x approaches 0 is equal to 1. We also know that the limit of 1/cos(x) as x approaches 0 is 1. Lastly, multiply the two together and you are left with the limit of tan(x)/x: 1.

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When dealing with a fraction within a fraction, there is a method of approaching it: keep, change, flip. In practice, you would find the common denominator of the fractions within the numerator, change the division sign to multiplication, flip the fractions in the denominator, and multiply.

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A continuous function is a function that can be drawn without picking up the pencil. The line is "continuous" if there are no disruptions or breaks.

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Continuity can also be determined at a point without evaluating/graphing the entire function. In order to do this, you must find the y value of the specified x in relation to the line of the function using limits.

There are three parts to finding continuity at a point:

1) The limit of the point from the left

2) The point itself (the y value when x is plugged in)

3) The limit of the point from the right

In order for there to be continuity at the point, all three parts must be equal to each other.

There are three parts to finding continuity at a point:

1) The limit of the point from the left

2) The point itself (the y value when x is plugged in)

3) The limit of the point from the right

In order for there to be continuity at the point, all three parts must be equal to each other.

f(x) = x³ - 3x² + 2x -6 where c = 2

The c simply indicates which x value you are using to evaluate the function. In this case, there is continuity as polynomial functions are always continuous.

To write this, we would say that

lim f(x) = f(2) = lim f(x)

x->2- x->2+

-6 = -6 = -6

There are three types of discontinuities with distinct features:

Asymptotes (unbound behavior) - All 3 parts are different ex. Limit of (x+6)/(x-6) as x approaches 6.

Jump - One of the two limits is equal to the point; the other limit is different

Hole - Both limits are the same; the point is different

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Piecewise functions are functions that are divided into parts and defined based on intervals within the domain. In order to find continuity within a piecewise function, you must find the limit from the left, the point, and the limit from the right as x approaches the c value given.

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There are two pieces: if we are to determine continuity using these two pieces, we must find the limit using the top piece while placing the y value at the defined location.

The c (not shown in picture) is equal to 1. Thus, we are finding the limit of the piecewise function as x approaches 1.

The function is a line (y=x+1) with the exception of the defined point (1, 3), indicating that there is a hole.

The c (not shown in picture) is equal to 1. Thus, we are finding the limit of the piecewise function as x approaches 1.

The function is a line (y=x+1) with the exception of the defined point (1, 3), indicating that there is a hole.

lim f(x) ≠ f(1) ≠ lim f(x)

x->1- x->1+

2 ≠ 3 ≠ 2

x->1- x->1+

2 ≠ 3 ≠ 2

An infinite limit is a limit in which f(x) increases and decreases without bound as x approaches c; this usually happens at a vertical asymptote. In order to find a vertical asymptote, the denominator must be set equal to 0 and solved. An example of this is attempting to find the limit of f((x+6)/(x-6) as x approaches 6—there would be a vertical asymptote at x=6 due to substitution resulting in a 0 in the denominator.

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The function 1/x also results in a vertical asymptote.

The primary method of finding horizontal asymptotes is BOBO BOTN EEDC, or Big on Bottom Zero, Big on Top None, Equal Exponents Divide Coefficients. In order to find horizontal asymptotes, the c must be positive or negative infinity to ensure BOBO is applicable.

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As the numerator's largest exponent is 1 and the denominator's largest exponent is 2, BOBO applies. In this case, as the function is Big On Bottom, the limit of f(x) as x approaches infinity is 0.

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As both the numerator and denominator's largest exponents are 2, the limit can be found by dividing the coefficients (EEDC). In this case, the limit of f(x) as x approaches infinity is 7/2.

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The limit of this function does not exist as it is heading toward both negative and positive infinity. The function has both a slant asymptote ((-2x/5)-(1/25)) and a vertical asymptote (3/5). As there was no horizontal asymptote, BOTN applied.

Limits to infinity are limits in which x is approaching negative or positive infinity. In these cases, the end behavior of the function is being evaluated.

Two general rules of thumb are that BOBO can be applied to limits toward infinity and 1/big (1 divided by infinity) is equal to 0.

Two general rules of thumb are that BOBO can be applied to limits toward infinity and 1/big (1 divided by infinity) is equal to 0.

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As BOBO applies, the coefficients can be divided. The limit is equal to -2/5.

First and foremost, left and right-hand limits, or one-sided limits, always exist on at least one side of the x. In order to find left and right-sided limits on a graph, you must find the c and then look at the left and the right to see what y is approaching from either side. In the function, finding the limit of f(x) as x approaches c- indicates that the c is being approached from the left whereas c+ indicates that c is being approached from the right.

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lim |x-2|/(x-2) = 1

x->2+

x->2+

lim |x-2|/(x-2) = -1

x->2-

x->2-

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lim f(x) = ∞

x->0+

x->0+

lim f(x) = -∞

x->0-

x->0-

The IVT, or Intermediate Value Theorem, says that if a function is continuous, there will always be a value between two Y values to match values between 2 X values. It does not indicate where the value is or what it is equal to; it simply indicates the existence of the values.

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Limits do not always exist in functions; if the limit from the left does not equal the limit from the right, the limit is DNE (does not exist). There are a few places where limits never exist:

1) Vertical asymptotes

2) Oscillating behavior

3) Piecewise functions (sometimes)

4) Vertical tangent

1) Vertical asymptotes

2) Oscillating behavior

3) Piecewise functions (sometimes)

4) Vertical tangent

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The limit of f(x) as x approaches 1 is DNE.

The limit of f(x) as x approaches 6 is DNE.

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