Limiting Euler

Today's question is about limits and Euler's number, e. This number has many interesting properties, and one of them is the derivative of e^x, which is e^x. Today's question does not deal with the derivative of this function, but with some limits.

Find the limit:
lim_{x -> infinity} (e^x + x)^1/x

This question is definitely not easy, mainly because of that annoying x in there and the 1/x exponent. However, we must never give up. We need to fight this question, so let's fight it with simplicity. Define another variable!

y = (e^x + x)^1/x

Simple, isn't it? All we have to do now is... well... find the limit of y. That didn't help. You can see there is a fraction in the exponent, so if we could just bring it down to a normal fraction, we could use L'Hopital's rule and solve this.

To get the exponent down, we can use a very well known technique: logs. There is a great rule in logarithms which states:
log_a bⁿ = nlog_a b

Using that fantastic rule we can bring the 1/x down and use L'Hopital's rule! To get a nice answer, let's take the logarithm of base e of this function, also written as ln (natural logarithm). So:
ln y = ln (e^x + x)^1/x

Using the logarithm rule mentioned earlier, we can say this equals:
ln (e^x + x) * 1/x
ln (e^x + x) / x

Mr. L'Hopital starts to smile. Now we can find the limit of this function and remember this is the logarithm of the function, so we'll need to consider that.

lim_{x -> infinity} ln y = lim_{x -> infinity} ln (e^x + x) / x

Using L'Hopital's rule (lim_{x -> a} f(x)/g(x) = lim_{x -> a} f'(x)/g'(x)), we can conclude that:
lim_{x -> infinity} ln (e^x + x) / x
is equal to:
lim_{x -> infinity} (e^x + 1) * (1 / (e^x + x)) / 1
lim_{x -> infinity} (e^x + 1) / (e^x + x)

Use L'Hopital's rule twice more:
lim_{x -> infinity} e^x / (e^x + 1)
lim_{x -> infinity} e^x / e^x

Now that looks familiar, doesn't it? This limit equals 1.

Remember: lim_{x -> infinity} ln y = 1, so lim_{x -> infinity} y = e

The limit of the entire function is e. Problem solved.

Hope you enjoyed,
Nadav

nadavs

Two Way Limits

It does not happen often, but when it does, it's simply beautiful: a question with two completely different yet right answers. I saw such question on Yahoo Answers today. Although I didn't answer it, the two different answers were so interesting and different in their approach, I had to bring it here.

The question is quite simple for people who know limits, but the two different answers are just amazing.

Find the limit:
lim_{x -> 4} (3 - (5 + x)^1/2) / (1 - (5 - x)^1/2)

As you can clearly see, plugging x = 4 gives a zero in the denominator, which is why the question asks for a limit. You can plug in numbers and get a rough estimation, but we want a definite answer.

The first method is using L'Hopital's rule. This rule says that in order to find lim_{x -> a} f(x)/g(x), you can also find lim_{x -> a} f'(x)/g'(x). All we need now is to differentiate the numerator and denominator and see what we get:
lim_{x -> 4} (-1/2 * (5 + x)^-1/2) / (1/2 * (5 - x)^-1/2)

The halves cancel, and by using the law that says aⁿ / bⁿ = (a/b)ⁿ, we can show that:
lim_{x -> 4} -((5 + x) / (5 - x))^-1/2

Now plug in 4 and you will get -((5 + 4)/(5-4))^-1/2
Which is -9^-1/2 = -1/3

This is one very good solution, and most people who know calculus would choose that. However, there is another solution to this problem. To do that, multiply the numerator and denominator by the conjugate of the denominator:
lim_{x -> 4} (3 - (5 + x)^1/2) / (1 - (5 - x)^1/2) * (1 + (5 - x)^1/2) / (1 + (5 - x)^1/2)

Using the law that says (a + b)(a - b) = a² - b² and the distributive property, we can conclude that:
lim_{x -> 4} (3 - (5 + x)^1/2) * (1 + (5 - x)^1/2) / (1 - (5 - x))
lim_{x -> 4} (3 - (5 + x)^1/2) * (1 + (5 - x)^1/2) / (x - 4)

Now multiply both parts of the fraction by the conjugate of the original numerator:
lim_{x -> 4} (3 - (5 + x)^1/2) * (1 + (5 - x)^1/2) / (x - 4) * (3 + (5 + x)^1/2) / (3 + (5 + x)^1/2)
lim_{x -> 4} (9 - (5 + x)) * (1 + (5 - x)^1/2) / ((x - 4) * (3 + (5 + x)^1/2))
lim_{x -> 4} (4 - x) * (1 + (5 - x)^1/2) / ((x - 4) * (3 + (5 + x)^1/2) )

That is really nice. (4 - x) / (x - 4) = -1, so we can cancel out two terms and turn them into a nice little minus sign:
lim_{x -> 4} -(1 + (5 - x)^1/2) / (3 + (5 + x)^1/2)

Now we can safely plug 4 for x and we get:
-(1 + (5 - 4)^1/2) / (3 + (5 + 4)^1/2)
-(1 + 1^1/2) / (3 + 9^1/2)
-(1 + 1) / (3 + 3)
-1/3

Again, we get -1/3 as an answer.

Math can be very easy if you think outside the box. Try doing it as often as you can.
Nadav

nadavs