Different ways of computing a limit (Part 2)

One of my colleagues placed the following problem on an exam for his Calculus II course…

\displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2+1}}{x}

and was impressed by the variety of correct responses that he received. I thought it would be fun to discuss some of the different ways that this limit can be computed.

Method #2. Using L’Hopital’s Rule. The limit has the indeterminant form \infty/\infty, and so I can differentiate the top and the bottom with respect to x:

\displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2+1}}{x} = \displaystyle \lim_{x \to \infty} \frac{ \displaystyle \frac{d}{dx} \left( \sqrt{x^2+1} \right) }{\displaystyle \frac{d}{dx} \left( x \right)}

= \displaystyle \lim_{x \to \infty} \frac{ \displaystyle \frac{1}{2} \left( x^2+1 \right)^{-1/2} \cdot 2x }{1}

= \displaystyle \lim_{x \to \infty} \frac{x}{\sqrt{x^2+1}}.

Oops… it looks like I just got the reciprocal of the original limit! Indeed, if I use L’Hopital’s Rule again, I’ll just return back to the original limit.

So that doesn’t look very helpful… except it is. If I define the value of this limit to be equal to L, then I’ve just shown that L = 1/L (assuming that the limit exists in the first place, of course). That means that L = 1 or L = -1. Well, clearly the limit of this nonnegative function can’t be negative, and so we conclude that the limit is equal to 1.

How I Impressed My Wife: Part 6a

This series was inspired by a question that my wife asked me: calculate

Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}

Originally, I multiplied the top and bottom of the integrand by \tan^2 x and performed a substitution. However, as I’ve discussed in this series, there are four different ways that this integral can be evaluated.
Starting with today’s post, I’ll begin a fifth method. I really like this integral, as it illustrates so many different techniques of integration as well as the trigonometric tricks necessary for computing some integrals.
green lineSince Q is independent of a, I can substitute any convenient value of a that I want without changing the value of Q. As shown in previous posts, substituting a =0 yields the following simplification:

Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}

= \displaystyle \int_{0}^{2\pi} \frac{dx}{\cos^2 x + 2 \cdot 0 \cdot \sin x \cos x + (0^2 + b^2) \sin^2 x}

= \displaystyle \int_{0}^{2\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}

= \displaystyle \int_{-\pi}^{\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}

= \displaystyle \int_{-\infty}^{\infty} \frac{ 2(1+u^2) du}{u^4 + (4 b^2 - 2) u^2 + 1}

Earlier, I evaluated this last integral using partial fractions, separating into the cases |b| = 1, |b| > 1, and |b| < 1. Now, I’ll calculate this same integral using contour integration. (See Wikipedia and Mathworld for more details.)

It turns out that Q can be rewritten as

Q = \displaystyle \lim_{R \to \infty} \oint_{C_R} \frac{ 2(1+z^2) dz}{z^4 + (4 b^2 - 2) z^2 + 1},

where C_R is the contour in the complex plane shown above (graphic courtesy of Mathworld). That’s because

\displaystyle \lim_{R \to \infty} \oint_{C_R} \frac{ 2(1+z^2) dz}{z^4 + (4 b^2 - 2) z^2 + 1}

= \displaystyle \lim_{R \to \infty} \int_{-R}^R \frac{ 2(1+z^2) dz}{z^4 + (4 b^2 - 2) z^2 + 1} + \lim_{R \to \infty} \int_{\gamma_R} \frac{ 2(1+z^2) dz}{z^4 + (4 b^2 - 2) z^2 + 1}

= \displaystyle \int_{-\infty}^{\infty} \frac{ 2(1+z^2) dz}{z^4 + (4 b^2 - 2) z^2 + 1} + \lim_{R \to \infty} \int_{\gamma_R} \frac{ 2(1+z^2) du}{z^4 + (4 b^2 - 2) z^2 + 1}

= Q + \displaystyle \lim_{R \to \infty} \int_{\gamma_R} \frac{ 2(1+z^2) dz}{z^4 + (4 b^2 - 2) z^2 + 1}

To show that the limit of the last integral is equal to 0, I use the parameterization z = R e^{i \theta}, so that dz = i R e^{i \theta}:

\displaystyle \lim_{R \to \infty} \left| \int_{\gamma_R} \frac{ 2(1+z^2) dz}{z^4 + (4 b^2 - 2) z^2 + 1} \right|

= \displaystyle \lim_{R \to \infty} \left| \int_0^{\pi} \frac{ 2R(1+R^2 e^{2i\theta}) d\theta}{R^4 e^{4 i\theta} + (4 b^2 - 2) R^2 e^{2i\theta} + 1} \right|

\le \displaystyle \lim_{R \to \infty} \pi \max_{0 \le \theta \le \pi} \left| \frac{ 2R(1+R^2 e^{2i\theta})}{R^4 e^{4 i\theta} + (4 b^2 - 2) R^2 e^{2i\theta} + 1} \right|

= \displaystyle \pi \max_{0 \le \theta \le \pi} \lim_{R \to \infty} \left| \frac{ 2R(1+R^2 e^{2i\theta})}{R^4 e^{4 i\theta} + (4 b^2 - 2) R^2 e^{2i\theta} + 1} \right|

= \displaystyle \pi \max_{0 \le \theta \le \pi} 0

= 0.

The above limit is equal to zero because the numerator grows like R^3 while the denominator grows like R^4. (This can be more laboriously established using L’Hopital’s rule).

Therefore, I have shown that

Q = \displaystyle \lim_{R \to \infty} \oint_{C_R} \frac{ 2(1+z^2) dz}{z^4 + (4 b^2 - 2) z^2 + 1},

and this contour integral can be computed using residues.

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I’ll continue with this fifth evaluation of the integral, starting with the case |b| = 1, in tomorrow’s post.

How I Impressed My Wife: Part 4h

So far in this series, I have used three different techniques to show that

Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x} = \displaystyle \frac{2\pi}{|b|}.

For the third technique, a key step in the calculation was showing that the residue of the function

f(z) = \displaystyle \frac{1}{z^2 + 2\frac{S}{R}z + 1} = \displaystyle \frac{1}{(z-r_1)(z-r_2)}

at the point

r_1 = \displaystyle \frac{-S + \sqrt{S^2 -R^2}}{R}

was equal to

\displaystyle \frac{R}{ 2 \sqrt{S^2-R^2} }.

Initially, I did this by explicitly computing the Laurent series expansion about z = r_1 and identifying the coefficient for the term (z-r_1)^{-1}.

In this post, I’d like to discuss another way that this residue could have been obtained.
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Notice that the function f(z) has the form \displaystyle \frac{g(z)}{(z-r) h(z)}, where g and h are differentiable functions so that g(r) \ne 0 and h(r) \ne 0. Therefore, we may rewrite this function using the Taylor series expansion of \displaystyle \frac{g(z)}{h(z)} about z = r:

f(z) = \displaystyle \frac{1}{z-r} \left[ \frac{g(z)}{h(z)} \right]

f(z) = \displaystyle \frac{1}{z-r} \left[ a_0 + a_1 (z-r) + a_2 (z-r)^2 + a_3 (z-r)^3 + \dots \right]

f(z) = \displaystyle \frac{a_0}{z-r} + a_1 + a_2 (z-r) + a_3 (z-r)^2 + \dots

Clearly,

\displaystyle \lim_{z \to r} (z-r) f(z) = \displaystyle \lim_{z \to r} \left[ a_0 + a_1 (z-r) + a_2 (z-r)^2 + a_3 (z-r)^3 + \dots \right] = a_0

Therefore, the residue at z = r can be found by evaluating the limit \displaystyle \lim_{z \to r} (z-r) f(z). Notice that

\displaystyle \lim_{z \to r} (z-r) f(z) = \displaystyle \lim_{z \to r} \frac{(z-r) g(z)}{(z-r) h(z)}

= \displaystyle \lim_{z \to r} \frac{(z-r) g(z)}{H(z)},

where H(z) = (z-r) h(z) is the original denominator of f(z). By L’Hopital’s rule,

a_0 = \displaystyle \lim_{z \to r} \frac{(z-r) g(z)}{H(z)} = \displaystyle \lim_{z \to r} \frac{g(z) + (z-r) g'(z)}{H'(z)} = \displaystyle \frac{g(r)}{H'(r)}.

For the function at hand, g(z) \equiv 1 and H(z) = z^2 + 2\frac{S}{R}z + 1, so that H'(z) = 2z + 2\frac{S}{R}. Therefore, the residue at z = r_1 is equal to

\displaystyle \frac{1}{2r_1+2 \frac{S}{R}} = \displaystyle \frac{1}{2 \displaystyle \frac{-S + \sqrt{S^2 -R^2}}{R} + 2 \frac{S}{R}}

= \displaystyle \frac{1}{ ~ 2 \displaystyle \frac{\sqrt{S^2 -R^2}}{R} ~ }

= \displaystyle \frac{R}{2 \sqrt{S^2-R^2}},

matching the result found earlier.

 

Different definitions of e (Part 10): Connecting the two definitions

In this series of posts, I consider how two different definitions of the number e are related to each other. The number e is usually introduced at two different places in the mathematics curriculum:

  1. Algebra II/Precalculus: If P dollars are invested at interest rate r for t years with continuous compound interest, then the amount of money after t years is A = Pe^{rt}.
  2. Calculus: The number e is defined to be the number so that the area under the curve y = 1/x from x = 1 to x = e is equal to 1, so that

\displaystyle \int_1^e \frac{dx}{x} = 1.

These two definitions appear to be very, very different. One deals with making money. The other deals with the area under a hyperbola. Amazingly, these two definitions are related to each other. In this series of posts, I’ll discuss the connection between the two.

logareagreen line

In yesterday’s post, I proved the following theorem, thus completing a long train of argument that began with the second definition of e and ending with a critical step in the derivation of the continuous compound interest formula. Today, I present an alternate proof of the theorem using L’Hopital’s rule.

Theorem. \displaystyle \lim_{n \to \infty} P \left( 1 + \frac{r}{n} \right)^{nt} = P e^{rt}.

Proof #2. Let’s write the left-hand side as

L = \displaystyle \lim_{n \to \infty} P \left( 1 + \frac{r}{n} \right)^{nt}.

Let’s take the natural logarithm of both sides:

\ln L = \displaystyle \ln \left[ \lim_{n \to \infty} P \left( 1 + \frac{r}{n} \right)^{nt} \right]

Since g(x) = \ln x is continuous, we can interchange the function and the limit on the right-hand side:

\ln L = \displaystyle \lim_{n \to \infty} \ln \left[ P \left( 1 + \frac{r}{n} \right)^{nt} \right]

\ln L = \displaystyle \lim_{n \to \infty} \left[ \ln P + \ln \left( 1 + \frac{r}{n} \right)^{nt} \right]

\ln L = \displaystyle \lim_{n \to \infty} \left[ \ln P + nt \ln \left( 1 + \frac{r}{n} \right)\right]

\ln L = \ln P + \displaystyle \lim_{n \to \infty} \frac{t \displaystyle \ln \left(1 + \frac{r}{n} \right)}{1/n}

\ln L = \ln P + \displaystyle t \lim_{n \to \infty} \frac{\displaystyle \ln \left(1 + \frac{r}{n} \right)}{1/n}

The limit on the right-hand side follows the indeterminate form 0/0, as so we may apply L’Hopital’s Rule. Taking the derivative of both the numerator and denominator with respect to n, we find

\ln L = \ln P + \displaystyle t \lim_{n \to \infty} \frac{\displaystyle \frac{1}{1 + \frac{r}{n}} \cdot \frac{-r}{n^2}}{\displaystyle \frac{-1}{n^2}}

\ln L = \ln P + \displaystyle t \lim_{n \to \infty}\frac{r}{1 + \frac{r}{n}}

\ln L = \ln P + \displaystyle t \frac{r}{1 + 0}

\ln L = rt + \ln P

We now solve for the original limit L:

L = e^{rt + \ln P}

L = e^{rt} e^{\ln P}

L = Pe^{rt}

 

Different definitions of e (Part 8): Connecting the two definitions

In this series of posts, I consider how two different definitions of the number e are related to each other. The number e is usually introduced at two different places in the mathematics curriculum:

  1. Algebra II/Precalculus: If P dollars are invested at interest rate r for t years with continuous compound interest, then the amount of money after t years is A = Pe^{rt}.
  2. Calculus: The number e is defined to be the number so that the area under the curve y = 1/x from x = 1 to x = e is equal to 1, so that

\displaystyle \int_1^e \frac{dx}{x} = 1.

These two definitions appear to be very, very different. One deals with making money. The other deals with the area under a hyperbola. Amazingly, these two definitions are related to each other. In this series of posts, I’ll discuss the connection between the two.

logareagreen line

In yesterday’s post, I proved the following theorem, thus completing a long train of argument that began with the second definition of e and ending with a critical step in the derivation of the continuous compound interest formula. Today, I present an alternate proof of the theorem using L’Hopital’s rule.

Theorem. \displaystyle \lim_{h \to 0} \left( 1 + h \right)^{1/h} = e.

Proof #2. Let’s write the left-hand side as

L = \displaystyle \lim_{h \to 0} \left( 1 + h \right)^{1/h}.

Let’s take the natural logarithm of both sides:

\ln L = \displaystyle \ln \left[ \lim_{h \to 0} \left( 1 + h \right)^{1/h} \right]

Since g(x) = \ln x is continuous, we can interchange the function and the limit on the right-hand side:

\ln L = \displaystyle \lim_{h \to 0} \ln (1+h)^{1/h}

\ln L = \displaystyle \lim_{h \to 0} \frac{1}{h} \ln(1+h)

\ln L = \displaystyle \lim_{h \to 0} \frac{\ln(1+h)}{h}

The right-hand side follows the indeterminate form 0/0, as so we may apply L’Hopital’s Rule. Taking the derivative of both the numerator and denominator, we find

\ln L = \displaystyle \lim_{h \to 0} \frac{ \displaystyle ~ \frac{1}{1+h} ~}{1}

\ln L = \displaystyle \lim_{h \to 0} \frac{1}{1+h}

\ln L = \displaystyle \frac{1}{1+0} = 1.

Therefore, the original limit is L = e^1 = e.

 

Why 0^0 is undefined

TI00

Here’s an explanation for why 0^0 is undefined that should be within the grasp of pre-algebra students:

Part 1.

  • What is 0^3? Of course, it’s 0.
  • What is 0^2? Again, 0.
  • What is 0^1? Again, 0.
  • What is 0^{1/2}, or \sqrt{0}? Again, 0.
  • What is 0^{1/3}, or \sqrt[3]{0}? In other words, what number, when cubed, is 0? Again, 0.
  • What is 0^{1/10}, or \sqrt[10]{0}? In other words, what number, when raised to the 10th power, is 0. Again, 0.

So as the exponent gets closer to 0, the answer remains 0. So, from this perspective, it looks like 0^0 ought to be equal to 0.

Part 2.

  • What is 3^0. Of course, it’s 1.
  • What is 2^0. Again, 1.
  • What is 1^0. Again, 1.
  • What is \left( \displaystyle \frac{1}{2} \right)^0? Again, 1
  • What is \left( \displaystyle \frac{1}{3} \right)^0. Again, 1
  • What is \left( \displaystyle \frac{1}{10} \right)^0? Again, 1

So as the base gets closer to 0, the answer remains 1. So, from this perspective, it looks like 0^0 ought to be equal to 1.

In conclusion: looking at it one way, 0^0 should be defined to be 0. From another perspective, 0^0 should be defined to be 1.

Of course, we can’t define a number to be two different things! So we’ll just say that 0^0 is undefined — just like dividing by 0 is undefined — rather than pretend that 0^0 switches between two different values.

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Here’s a more technical explanation about why 0^0 is an indeterminate form, using calculus.

Part 1. As before,

\displaystyle \lim_{x \to 0^+} 0^x = \lim_{x \to 0^+} 0 = 0.

The first equality is true because, inside of the limit, x is permitted to get close to 0 but cannot actually equal 0, and there’s no ambiguity about 0^x = 0 if x >0. (Naturally, 0^x is undefined if x < 0.)

The second equality is true because the limit of a constant is the constant.

Part 2. As before,

\displaystyle \lim_{x \to 0} x^0 = \lim_{x \to 0} 1 = 1.

Once again, the first equality is true because, inside of the limit, x is permitted to get close to 0 but cannot actually equal 0, and there’s no ambiguity about x^0 = 1 if x \ne 0.

As before, the answers from Parts 1 and 2 are different. But wait, there’s more…

Part 3. Here’s another way that 0^0 can be considered, just to give us a headache. Let’s evaluate

\displaystyle \lim_{x \to 0^+} x^{1/\ln x}

Clearly, the base tends to 0 as x \to 0. Also, \ln x \to \infty as x \to 0^+, so that \displaystyle \frac{1}{\ln x} \to 0 as x \to 0^+. In other words, this limit has the indeterminate form 0^0.

To evaluate this limit, let’s take a logarithm under the limit:

\displaystyle \lim_{x \to 0^+} \ln x^{1/\ln x} = \displaystyle \lim_{x \to 0^+} \frac{1}{\ln x} \cdot \ln x

\displaystyle \lim_{x \to 0^+} \ln x^{1/\ln x} = \displaystyle \lim_{x \to 0^+} 1

\displaystyle \lim_{x \to 0^+} \ln x^{1/\ln x} = 1

Therefore, without the extra logarithm,

\displaystyle \lim_{x \to 0^+} x^{1/\ln x} = e^1 = e

Part 4. It gets even better. Let k be any positive real number. By the same logic as above,

\displaystyle \lim_{x \to 0^+} x^{\ln k/\ln x} = e^{\ln k} = k

So, for any k \ge 0, we can find a function f(x) of the indeterminate form 0^0 so that \displaystyle f(x) = k.

In other words, we could justify defining 0^0 to be any nonnegative number. Clearly, it’s better instead to simply say that 0^0 is undefined.

P.S. I don’t know if it’s possible to have an indeterminate form of 0^0 where the answer is either negative or infinite. I tend to doubt it, but I’m not sure.

A surprising appearance of e

Here’s a simple probability problem that should be accessible to high school students who have learned the Multiplication Rule:

Suppose that you play the lottery every day for about 20 years. Each time you play, the chance that you win is 1 chance in 1000. What is the probability that, after playing  1000 times, you never win?

This is a straightforward application of the Multiplication Rule from probability. The chance of not winning on any one play is 0.999. Therefore, the chance of not winning 1000 consecutive times is (0.999)^{1000}, which we can approximate with a calculator.

TIlottery1

Well, that was easy enough. Now, just for the fun of it, let’s find the reciprocal of this answer.

TIlottery2

Hmmm. Two point seven one. Where have I seen that before? Hmmm… Nah, it couldn’t be that.

What if we changed the number 1000 in the above problem to 1,000,000? Then the probability would be (0.999999)^{1000000}.

TIlottery3

There’s no denying it now… it looks like the reciprocal is approximately e, so that the probability of never winning for both problems is approximately 1/e.

Why is this happening? I offer a thought bubble if you’d like to think about this before proceeding to the answer.

green_speech_bubbleThe above calculations are numerical examples that demonstrate the limit

\displaystyle \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n = e^x

In particular, for the special case when n = -1, we find

\displaystyle \lim_{n \to \infty} \left(1 - \frac{1}{n}\right)^n = e^{-1} = \displaystyle \frac{1}{e}

The first limit can be proved using L’Hopital’s Rule. By continuity of the function f(x) = \ln x, we have

\ln \left[ \displaystyle \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n \right] = \displaystyle \lim_{n \to \infty} \ln \left[ \left(1 + \frac{x}{n}\right)^n \right]

\ln \left[ \displaystyle \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n \right] = \displaystyle \lim_{n \to \infty} n \ln \left(1 + \frac{x}{n}\right)

\ln \left[ \displaystyle \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n \right] = \displaystyle \lim_{n \to \infty} \frac{ \displaystyle \ln \left(1 + \frac{x}{n}\right)}{\displaystyle \frac{1}{n}}

The right-hand side has the form \infty/\infty as n \to \infty, and so we may use L’Hopital’s rule, differentiating both the numerator and the denominator with respect to n.

\ln \left[ \displaystyle \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n \right] = \displaystyle \lim_{n \to \infty} \frac{ \displaystyle \frac{1}{1 + \frac{x}{n}} \cdot \frac{-x}{n^2} }{\displaystyle \frac{-1}{n^2}}

\ln \left[ \displaystyle \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n \right] = \displaystyle \lim_{n \to \infty} \displaystyle \frac{x}{1 + \frac{x}{n}}

\ln \left[ \displaystyle \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n \right] = \displaystyle \frac{x}{1 + 0}

\ln \left[ \displaystyle \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n \right] = x

Applying the exponential function to both sides, we conclude that

\displaystyle \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n= e^x

green lineIn an undergraduate probability class, the problem can be viewed as a special case of a Poisson distribution approximating a binomial distribution if there’s a large number of trials and a small probability of success.

The above calculation also justifies (in Algebra II and Precalculus) how the formula for continuous compound interest A = Pe^{rt} can be derived from the formula for discrete compound interest A = P \displaystyle \left( 1 + \frac{r}{n} \right)^{nt}

All this to say, Euler knew what he was doing when he decided that e was so important that it deserved to be named.