Confusing limit question (math)

Problem: The limit as x approaches infinity of (3x^2+2x)/(x+5)-3x

The actual limit seems to be -13. But below are two explanations for different limits. I know they are wrong, but I can’t figure out what exactly is wrong with them. Here they are:

The limit is 0. X is very large, so 2x is small compared to 3x^2, and 5 is small in comparison to x. So, (3x^2+2x)/(x+5) behaves like 3x^2/x=3x. When we plug this in to the original limit we get 3x-3x, which equals 0. Thus the limit is 0.

The limit is 2. (3x^2+2x)/(x+5)=(3x+2)/(1+5/x), and X is a very large number. Thus 5/x is essentially 0, and so the denominator is essentially 1. Dividing (3x+2) by 1, and placing this all back into the original limit, we get 3x+2-3x, which equals 2. Thus the limit is 2.

Obviously there is something wrong with both of these explanations, but I cant figure out what. It’s driving me nuts! Any explanations for why these are wrong?

The thing is, when you are figuring out limits, you can’t just add, subtract, and divide infinities as if they were numbers. It seems to me that this is what you are doing.

For example, what is the limit as x approaches infinity of

3x - (3x + 3). Clearly the limit is 3, but by your approach, the limit would be zero. Think about it.

rewrite it over a single (x+5) denominator:

= (3x^2+2x)/(x+5) - (3x^2 + 15x)/(x+5)

= -13x/(x+5)

The problem with your explanations is in using weasel words like “behaves like” and “essentially” to ignore parts of the expression and assuming you can then just substitute the simplified forms into the original, and draw a valid conclusion.

In fact, the point of the problem is that the first term behaves very damned much like 3x as it approaches infinity, and the limit when 3x is subtracted is going to be very dependent on the rest of the expression.

Taking a break from my damn take-home partial differential equations test to post. Ugh, can’t take it anymore.

In situations like this, the first thing you should think of, or at least I did, is L’Hospital’s Rule. Not really so present in this case, but the main thing that you need to do is reduce it to one fraction. It’s bad juju to do things like subtracting infinities. Dividing infinities is fine by the above rule, but you have to make sure things work out.

So in your case, you change your function:

(3x^2 + 2x) / (x+5) - 3x

Multiply the 3x by (x+5) / (x+5) to get it into the same denominator, which gets you:

((3x^2 + 2x ) - (3x^2 + 15x)) / (x+5)

Simplifying, you get:

(-13x) / (x+5)

Now, you can either use L’Hospital, like I usually would since it’s easier, IMHO, or multiply by (1/x) / (1/x):

(-13) / (1 + 5/x)

Now, you know the limits of -13, 1, and 5/x as they go to inf. are -13, 1, and 0 respectively, so we end up with:

(-13) / (1 + 0)

Or -13, as you stated.

And now, on preview, I’ve been beated by a correct answer. But as least I was long winded.

It is somewhat rare that L’Hospital’s rule is actually useful. It often replaces understanding with mindless computation.

Once the expression is written as a single rational expression, if you are taking the limit at infinity, you can ignore the lower order terms. (This is often an exercise in calc books.) So once it is written as

(-13x)/(x+5),

the limit is the limit of (-13x)/(x) which is -13.

Jay

The quick-and-dirty physicist’s way of doing it:

(3x[sup]2[/sup] + 2x)/(x + 5) - 3x
= (3x[sup]2[/sup] + 2x) x[sup]-1[/sup] (1 + 5/x)[sup]-1[/sup] - 3x
= (3x + 2) (1 - 5/x + 25/x[sup]2[/sup] - …) - 3x
= (3x + (2 - 15) + (75 - 10)/x + … ) - 3x
= -13 + 65/x + …

which obviously approaches -13 as x increases without bound. The dodgy step here is the expansion of (1 + 5/x)[sup]-1[/sup] in a power series in (5/x), in the third step; I wouldn’t recommend pulling this kind of stunt in a first-year calculus course, but if you go over to the physics building this kind of method is used all the time.

(Of course, us physicists also believe that the sum of the integers from 1 to infinity is -1/12. So be careful.)

Yes, I have to admit, in my past, I preferred ‘mindless computation’ instead of actual understanding, as the majority of my math knowledge was directed towards Mu Alpha Theta competitions instead of true understanding.

Since then, it’s come much easier to me to just think things through using L’Hospital instead of looking at the primary terms since that what I was trained to do.

Yeah, my comment came out different than I intended. L’Hospital’s rule is often useful, it isn’t often necessary. And mindless calculation, of course, isn’t always a bad thing. I am reminded of the following quote:

Jay