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In [link] we look at one-sided limits of a piecewise-defined function and use these limits to draw a conclusion about a two-sided limit of the same function.

Evaluating a two-sided limit using the limit laws

For f ( x ) = { 4 x 3 if x < 2 ( x 3 ) 2 if x 2 , evaluate each of the following limits:

  1. lim x 2 f ( x )
  2. lim x 2 + f ( x )
  3. lim x 2 f ( x )

[link] illustrates the function f ( x ) and aids in our understanding of these limits.

The graph of a piecewise function with two segments. For x<2, the function is linear with the equation 4x-3. There is an open circle at (2,5). The second segment is a parabola and exists for x>=2, with the equation (x-3)^2. There is a closed circle at (2,1). The vertex of the parabola is at (3,0).
This graph shows a function f ( x ) .
  1. Since f ( x ) = 4 x 3 for all x in ( , 2 ) , replace f ( x ) in the limit with 4 x 3 and apply the limit laws:
    lim x 2 f ( x ) = lim x 2 ( 4 x 3 ) = 5 .
  2. Since f ( x ) = ( x 3 ) 2 for all x in ( 2 , + ) , replace f ( x ) in the limit with ( x 3 ) 2 and apply the limit laws:
    lim x 2 + f ( x ) = lim x 2 ( x 3 ) 2 = 1 .
  3. Since lim x 2 f ( x ) = 5 and lim x 2 + f ( x ) = 1 , we conclude that lim x 2 f ( x ) does not exist.
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Graph f ( x ) = { x 2 if x < 1 2 if x = −1 x 3 if x > 1 and evaluate lim x −1 f ( x ) .


The graph of a piecewise function with three segments. The first is a linear function, -x-2, for x<-1. The x intercept is at (-2,0), and there is an open circle at (-1,-1). The next segment is simply the point (-1, 2). The third segment is the function x^3 for x > -1, which crossed the x axis and y axis at the origin.
lim x −1 f ( x ) = −1

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We now turn our attention to evaluating a limit of the form lim x a f ( x ) g ( x ) , where lim x a f ( x ) = K , where K 0 and lim x a g ( x ) = 0 . That is, f ( x ) / g ( x ) has the form K / 0 , K 0 at a .

Evaluating a limit of the form K / 0 , K 0 Using the limit laws

Evaluate lim x 2 x 3 x 2 2 x .

Step 1. After substituting in x = 2 , we see that this limit has the form −1 / 0 . That is, as x approaches 2 from the left, the numerator approaches −1; and the denominator approaches 0. Consequently, the magnitude of x 3 x ( x 2 ) becomes infinite. To get a better idea of what the limit is, we need to factor the denominator:

lim x 2 x 3 x 2 2 x = lim x 2 x 3 x ( x 2 ) .

Step 2. Since x 2 is the only part of the denominator that is zero when 2 is substituted, we then separate 1 / ( x 2 ) from the rest of the function:

= lim x 2 x 3 x · 1 x 2 .

Step 3. lim x 2 x 3 x = 1 2 and lim x 2 1 x 2 = . Therefore, the product of ( x 3 ) / x and 1 / ( x 2 ) has a limit of +∞:

lim x 2 x 3 x 2 2 x = + .
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Evaluate lim x 1 x + 2 ( x 1 ) 2 .

+∞

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The squeeze theorem

The techniques we have developed thus far work very well for algebraic functions, but we are still unable to evaluate limits of very basic trigonometric functions. The next theorem, called the squeeze theorem    , proves very useful for establishing basic trigonometric limits. This theorem allows us to calculate limits by “squeezing” a function, with a limit at a point a that is unknown, between two functions having a common known limit at a . [link] illustrates this idea.

A graph of three functions over a small interval. All three functions curve. Over this interval, the function g(x) is trapped between the functions h(x), which gives greater y values for the same x values, and f(x), which gives smaller y values for the same x values. The functions all approach the same limit when x=a.
The Squeeze Theorem applies when f ( x ) g ( x ) h ( x ) and lim x a f ( x ) = lim x a h ( x ) .

The squeeze theorem

Let f ( x ) , g ( x ) , and h ( x ) be defined for all x a over an open interval containing a . If

f ( x ) g ( x ) h ( x )

for all x a in an open interval containing a and

lim x a f ( x ) = L = lim x a h ( x )

where L is a real number, then lim x a g ( x ) = L .

Applying the squeeze theorem

Apply the squeeze theorem to evaluate lim x 0 x cos x .

Because −1 cos x 1 for all x , we have x x cos x x for x 0 and x x c o s x x for x 0 (if x is negative the direction of the inequalities changes when we multiply). Since lim x 0 ( x ) = 0 = lim x 0 x , from the squeeze theorem, we obtain lim x 0 x cos x = 0 . The graphs of f ( x ) = x , g ( x ) = x cos x , and h ( x ) = x are shown in [link] .

The graph of three functions: h(x) = x, f(x) = -x, and g(x) = xcos(x). The first, h(x) = x, is a linear function with slope of 1 going through the origin. The second, f(x), is also a linear function with slope of −1; going through the origin. The third, g(x) = xcos(x), curves between the two and goes through the origin. It opens upward for x>0 and downward for x>0.
The graphs of f ( x ) , g ( x ) , and h ( x ) are shown around the point x = 0 .
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Use the squeeze theorem to evaluate lim x 0 x 2 sin 1 x .

0

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We now use the squeeze theorem to tackle several very important limits. Although this discussion is somewhat lengthy, these limits prove invaluable for the development of the material in both the next section and the next chapter. The first of these limits is lim θ 0 sin θ . Consider the unit circle shown in [link] . In the figure, we see that sin θ is the y -coordinate on the unit circle and it corresponds to the line segment shown in blue. The radian measure of angle θ is the length of the arc it subtends on the unit circle. Therefore, we see that for 0 < θ < π 2 , 0 < sin θ < θ .

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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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