diff --git a/source/chap-3.xml b/source/chap-3.xml index 8ab5102..e17c230 100644 --- a/source/chap-3.xml +++ b/source/chap-3.xml @@ -4,7 +4,7 @@ Using Derivatives - + diff --git a/source/sec-3-2-LHR.xml b/source/sec-3-2-LHR.xml index 8eb17ca..3e37acd 100644 --- a/source/sec-3-2-LHR.xml +++ b/source/sec-3-2-LHR.xml @@ -3,7 +3,7 @@
Using derivatives to evaluate limits - + - Introduction + + Foundations

- Because differential calculus is based on the definition of the derivative, - and the definition of the derivative involves a limit, - there is a sense in which all of calculus rests on limits. - In addition, - the limit involved in the definition of the derivative always generates the indeterminate form \frac{0}{0}. - If f is a differentiable function, - then in the definition - - f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} - , - not only does h \to 0 in the denominator, - but also (f(x+h)-f(x)) \to 0 in the numerator, - since f is continuous. - Remember, saying that a limit has an indeterminate form only means that we don't yet know its value and have more work to do: - indeed, limits of the form - \frac{0}{0} can take on any value, - as is evidenced by evaluating f'(x) for varying values of x for a function such as f(x) = x^2. -

- -

- We have learned many techniques for evaluating the limits that result from the derivative definition, - including a large number of shortcut rules. - In this section, we turn the situation upside-down: - instead of using limits to evaluate derivatives, - we explore how to use derivatives to evaluate certain limits. -

- - - - - - - - Using derivatives to evaluate indeterminate limits of the form <m>\frac{0}{0}</m>. -
- At left, the graphs of f and g near the value a, along with their tangent line approximations L_f and L_g at x = a. At right, zooming in on the point a and the four graphs. - -
- -

- The idea demonstrated in Preview Activity - that we can evaluate an indeterminate limit of the form - \frac{0}{0} by replacing each of the numerator and denominator with their local linearizations at the point of interest can be generalized in a way that enables us to evaluate a wide range of limits. - We have a function h(x) that can be written as a quotient - h(x) = \frac{f(x)}{g(x)}, - where f and g are both differentiable at x=a and for which f(a) = g(a) = 0. - We would like to evaluate the indeterminate limit given by - \lim_{x \to a} h(x). - Figure illustrates the situation. - We see that both f and g have an x-intercept at x = a. - Their respective tangent line approximations L_f and L_g at x = a - are also shown in the figure. - We can take advantage of the fact that a function and its tangent line approximation become indistinguishable as x \to a. -

- -

- First, let's recall that L_f(x) = f'(a)(x-a) + f(a) and L_g(x) = g'(a)(x-a) +g(a). - Because x is getting arbitrarily close to a when we take the limit, - we can replace f with L_f and replace g with L_g, - and thus we observe that - - \lim_{x \to a} \frac{f(x)}{g(x)} \amp= \lim_{x \to a} \frac{L_f(x)}{L_g(x)} - \amp= \lim_{x \to a} \frac{f'(a)(x-a) + f(a)}{g'(a)(x-a) + g(a)} - . -

- -

- Next, we remember that both f(a) = 0 and g(a) = 0, - which is precisely what makes the original limit indeterminate. - Substituting these values for f(a) and g(a) in the limit above, - we now have - - \lim_{x \to a} \frac{f(x)}{g(x)} \amp= \lim_{x \to a} \frac{f'(a)(x-a)}{g'(a)(x-a)} - \amp= \lim_{x \to a} \frac{f'(a)}{g'(a)} - , - where the latter equality holds because - \frac{x-a}{x-a} = 1 when x is approaching - (but not equal to) a. - Finally, we note that \frac{f'(a)}{g'(a)} is constant with respect to x, and thus - - \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{f'(a)}{g'(a)} - . -

- -

- This result holds as long as g'(a) is not equal to zero. - The formal name of the result is L'Hôpital's Rule. -

- - - L'Hôpital's Rule - L'Hôpital's rule - -

- Let f and g be differentiable on an open interval that includes x=a, - and suppose that f(a) = g(a) = 0 and that g'(a) \neq 0. - Then - - \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{f'(a)}{g'(a)}. - -

- - -

- In practice, - we typically work with a slightly more general version of L'Hôpital's Rule, - which states that (under the identical assumptions as the boxed rule above plus the extra assumption that g' is continuous at x=a) - - \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} - , - provided the righthand limit exists. - This form reflects the basic idea of L'Hôpital's Rule: - if \frac{f(x)}{g(x)} produces an indeterminate limit of form \frac{0}{0} as x \to a, - that limit is equivalent to the limit of the quotient of the two functions' derivatives, - \frac{f'(x)}{g'(x)}. -

- -

- For example, - if we consider the limit from Preview Activity, - - \lim_{x \to 1} \frac{x^5 + x - 2}{x^2 - 1} - , - by L'Hôpital's Rule we have that - - \lim_{x \to 1} \frac{x^5 + x - 2}{x^2 - 1} = \lim_{x \to 1} \frac{5x^4 + 1}{2x} = \frac{6}{2} = 3 - . -

- -

- By replacing the numerator and denominator with their respective derivatives, - we often replace an indeterminate limit with one whose value we can easily determine. -

- - - -

- While L'Hôpital's Rule can be applied in an entirely algebraic way, - it is important to remember that the justification of the rule is graphical: - the main idea is that the slopes of the tangent lines to f and g at x = a determine the value of the limit of \frac{f(x)}{g(x)} as x \to a. -

- -
- Two functions f and g that satisfy L'Hôpital's Rule. - -
- -

- We see this in Figure, - where we can see from the grid that - f'(a) = 2 and g'(a) = -1, - hence by L'Hôpital's Rule, - - \lim_{x \to a}\frac{f(x)}{g(x)} = \frac{f'(a)}{g'(a)} = \frac{2}{-1} = -2 - . -

- -

- It's not the fact that f and g both approach zero that matters most, - but rather the rate - at which each approaches zero that determines the value of the limit. - This is a good way to remember what L'Hôpital's Rule says: - if f(a) = g(a) = 0, - the the limit of \frac{f(x)}{g(x)} as x \to a is given by the ratio of the slopes of f and g at x = a. -

- - - - - - Limits involving <m>\infty</m> -

- The concept of infinity, - infinity - denoted \infty, arises naturally in calculus, - as it does in much of mathematics. - It is important to note from the outset that \infty is a concept, - but not a number itself. - Indeed, the notion of \infty naturally invokes the idea of limits. - Consider, for example, the function f(x) = \frac{1}{x}, - whose graph is pictured in Figure. -

- -

- We note that x = 0 is not in the domain of f, - so we may naturally wonder what happens as x \to 0. - As x \to 0^+, we observe that f(x) - increases without bound. - That is, we can make the value of f(x) as large as we like by taking x closer and closer - (but not equal) - to 0, while keeping x \gt 0. - This is a good way to think about what infinity represents: - a quantity is tending to infinity if there is no single number that the quantity is always less than. -

- -
- The graph of f(x) = \frac{1}{x}. - -
- -

- Recall that the statement \lim_{x \to a} f(x) = L, - means that we can make f(x) as close to L as we'd like by taking x sufficiently close - (but not equal) - to a. - We now expand this notation and language to include the possibility that either L or a can be \infty. - For instance, for f(x) = \frac{1}{x}, we now write - - \lim_{x \to 0^+} \frac{1}{x} = \infty - , - by which we mean that we can make - \frac{1}{x} as large as we like by taking x sufficiently close - (but not equal) - to 0. - In a similar way, we write - - \lim_{x \to \infty} \frac{1}{x} = 0 - , - since we can make \frac{1}{x} as close to 0 as we'd like by taking x sufficiently large (i.e., by letting x increase without bound). -

- -

- In general, the notation - \lim_{x \to a} f(x) = \infty means that we can make f(x) as large as we like by taking x sufficiently close - (but not equal) - to a, and the notation - \lim_{x \to \infty} f(x) = L means that we can make f(x) as close to L as we like by taking x sufficiently large. - This notation also applies to left- and right-hand limits, - and to limits involving -\infty. - For example, returning to Figure - and f(x) = \frac{1}{x}, we can say that - - \lim_{x \to 0^-} \frac{1}{x} = -\infty \ \ \text{and} \ \ \lim_{x \to -\infty} \frac{1}{x} = 0 - . -

- -

- Finally, we write - - \lim_{x \to \infty} f(x) = \infty - - if we can make the value of f(x) as large as we'd like by taking x sufficiently large. - For example, - - \lim_{x \to \infty} x^2 = \infty - . -

- -

- Limits involving infinity identify vertical - and horizontal asymptotes - asymptote - asymptotevertical - asymptotehorizontal - of a function. - If \lim_{x \to a} f(x) = \infty, - then x = a is a vertical asymptote of f, - while if \lim_{x \to \infty} f(x) = L, - then y = L is a horizontal asymptote of f. - Similar statements can be made using -\infty, - and with left- and right-hand limits as x \to a^- or x \to a^+. -

- -

- In precalculus classes, - it is common to study the end behavior - of certain families of functions, - by which we mean the behavior of a function as - x \to \infty and as x \to -\infty. - Here we briefly examine some familiar functions and note the values of several limits involving \infty. -

- -
- Graphs of some familiar functions whose end behavior as x \to \pm \infty is known. In the middle graph, f(x) = x^3 - 16x and g(x) = x^4 - 16x^2 - 8. - -
- -

- For the natural exponential function e^x, - we note that \lim_{x \to \infty} e^x = \infty and \lim_{x \to -\infty} e^x = 0. - For the exponential decay function e^{-x}, - these limits are reversed, - with \lim_{x \to \infty} e^{-x} = 0 and \lim_{x \to -\infty} e^{-x} = \infty. - Turning to the natural logarithm function, - we have \lim_{x \to 0^+} \ln(x) = -\infty and \lim_{x \to \infty} \ln(x) = \infty. - While both e^x and \ln(x) grow without bound as x \to \infty, - the exponential function does so much more quickly than the logarithm function does. - We'll soon use limits to quantify what we mean by quickly. -

- -

- For polynomial functions of the form - - p(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0 - , - the end behavior depends on the sign of a_n and whether the highest power n is even or odd. - If n is even and a_n is positive, - then \lim_{x \to \infty} p(x) = \infty and \lim_{x \to -\infty} p(x) = \infty, - as in the plot of g in Figure. - If instead a_n is negative, - then \lim_{x \to \infty} p(x) = -\infty and \lim_{x \to -\infty} p(x) = -\infty. - In the situation where n is odd, - then either \lim_{x \to \infty} p(x) = \infty and \lim_{x \to -\infty} p(x) = -\infty - (which occurs when a_n is positive, - as in the graph of f in Figure), - or \lim_{x \to \infty} p(x) = -\infty and - \lim_{x \to -\infty} p(x) = \infty - (when a_n is negative). -

- -

- A function can fail to have a limit as x \to \infty. - For example, - consider the plot of the sine function at right in Figure. - Because the function continues oscillating between -1 and 1 as x \to \infty, - we say that \lim_{x \to \infty} \sin(x) does not exist. -

- -

- Finally, it is straightforward to analyze the behavior of any rational function as x \to \infty. -

- - - -

- Determine the limit of the function - - q(x) = \frac{3x^2 - 4x + 5}{7x^2 + 9x - 10} - - as x \to \infty. -

- - -

- Note that both (3x^2 - 4x + 5) \to \infty as x \to \infty and - (7x^2 + 9x - 10) \to \infty as x \to \infty. - Here we say that \lim_{x \to \infty} q(x) has indeterminate form \frac{\infty}{\infty}. - We can determine the value of this limit through a standard algebraic approach. - Multiplying the numerator and denominator each by \frac{1}{x^2} (we choose this quantity since x^2 is the highest power present in the numerator or denominator), - we find that - - \lim_{x \to \infty} q(x) \amp= \lim_{x \to \infty} \frac{3x^2 - 4x + 5}{7x^2 + 9x - 10} \cdot \frac{\frac{1}{x^2}}{\frac{1}{x^2}} - \amp= \lim_{x \to \infty} \frac{3 - 4\frac{1}{x} + 5\frac{1}{x^2}}{7 + 9\frac{1}{x} - 10\frac{1}{x^2}} = \frac{3}{7} - - since \frac{1}{x^2} \to 0 and - \frac{1}{x} \to 0 as x \to \infty. - This shows that the rational function q has a horizontal asymptote at y = \frac{3}{7}. - A similar approach can be used to determine the limit of any rational function as x \to \infty. -

- - - -

- But how should we handle a limit such as - - \lim_{x \to \infty} \frac{x^2}{e^x}? - -

- -

- Here, both x^2 \to \infty and e^x \to \infty. Like the situation we considered in Example, \lim_{x \to \infty} \frac{3x^2 - 4x + 5}{7x^2 + 9x - 10}, here we have both x^2 \to \infty and e^x \to \infty, so - - \lim_{x \to \infty} \frac{x^2}{e^x} - - is also indeterminate with form \frac{\infty}{\infty}. - With the presence of e^x, there is not an algebraic approach that enables us to find the limit's value. - Fortunately, - it turns out that L'Hôpital's Rule can be extendedThe mathematical argument for why this is possible is beyond the scope of this text. One reason to think that an infinite version of L'Hôpital's Rule is plausible is that if we have \frac{f(x)}{g(x)} \to \frac{\infty}{\infty}, it's equivalent to consider \frac{1/g(x)}{1/f(x)} which has the indeterminate form \frac{0}{0}. to cases involving infinity. -

- - - L'Hôpital's Rule (<m>\infty</m>) - L'Hôpital's rule (at infinity) - -

- Suppose that f and g are both differentiable on an open interval that contains a, both approach zero or both approach \pm \infty as x \to a, and g'(x) \ne 0 on an open interval that contains a - (where a is allowed to be \infty), then - - \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} - , - provided the righthand limit exists. -

- - - - -

- To evaluate \lim_{x \to \infty} \frac{x^2}{e^x}, - we can now apply the infinite version of L'Hôpital's Rule, - since both x^2 \to \infty and e^x \to \infty. - Doing so, it follows that - - \lim_{x \to \infty} \frac{x^2}{e^x} = \lim_{x \to \infty} \frac{2x}{e^x} - . -

- -

- This updated limit is still indeterminate and of the form \frac{\infty}{\infty}, - but it is simpler since 2x has replaced x^2. - Hence, we can apply L'Hôpital's Rule again, and find that - - \lim_{x \to \infty} \frac{x^2}{e^x} = \lim_{x \to \infty} \frac{2x}{e^x} = \lim_{x \to \infty} \frac{2}{e^x} - . -

- -

- Now, since 2 is constant and - e^x \to \infty as x \to \infty, - it follows that \frac{2}{e^x} \to 0 as x \to \infty, - which shows that - - \lim_{x \to \infty} \frac{x^2}{e^x} = 0 - . -

- - - -

- To evaluate the limit of a quotient of two functions - \frac{f(x)}{g(x)} that results in an indeterminate form of \frac{\infty}{\infty}, - in essence we are asking which function is growing faster without bound. - We say that the function g dominates - the function f as x \to \infty provided that - - \lim_{x \to \infty} \frac{f(x)}{g(x)} = 0 - , - whereas f dominates g provided that \lim_{x \to \infty} \frac{f(x)}{g(x)} = \infty. - Finally, if the value of \lim_{x \to \infty} \frac{f(x)}{g(x)} is finite and nonzero, - we say that f and g - grow at the same rate. - For example, - we saw that \lim_{x \to \infty} \frac{x^2}{e^x} = 0, - so e^x dominates x^2 (i.e., e^x grows faster than x^2 as x \to \infty), - while \lim_{x \to \infty} \frac{3x^2 - 4x + 5}{7x^2 + 9x - 10} = \frac{3}{7}, - so f(x) = 3x^2 - 4x + 5 and - g(x) = 7x^2 + 9x - 10 grow at the same rate. -

- - - - Summary -

-

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  • -

    - Derivatives can be used to help us evaluate indeterminate limits of the form - \frac{0}{0} through L'Hôpital's Rule, - by replacing the functions in the numerator and denominator with their tangent line approximations. - In particular, - if f(a) = g(a) = 0 and f and g are differentiable on an open interval containing a, L'Hôpital's Rule tells us that - - \lim_{x \to a}\frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f'(x)}{g'(x)} - . -

    -
    • - -
  • -

    - When we write x \to \infty, - this means that x is increasing without bound. - Thus, writing \lim_{x \to \infty} f(x) = L - means that we can make f(x) as close to L as we like by choosing x to be sufficiently large. - Similarly, \lim_{x \to a} f(x) = \infty - means that we can make f(x) as large as we like by choosing x sufficiently close to a. -

    -
    • - -
  • -

    - A version of L'Hôpital's Rule also helps us evaluate indeterminate limits of the form \frac{\infty}{\infty}. - If f and g are differentiable and both approach zero or both approach \pm \infty as x \to a - (where a is allowed to be \infty), - then - - \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} - . -

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+ Coming soon.

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