Limit Exercise 1

[DrToddMatthews]

I thought I was done working with limits, but I was wrong.

Assuming that the limits for f(x) and g(x) as x->a exist, then the following
is a property of limits:

    lim  f(x)
    x->a             lim  [f(x)/g(x)]    where  lim g(x)  does not equal zero
  -------------  =   x->a                       x->a
    lim  g(x)
    x->a

The exercise is a TRUE or FALSE question.

Given that:

    lim  f(x) = 0    and     lim  g(x) = 0
    x->5                     x->5


then:


    lim  [f(x)/g(x)]  does not exist
    x->5

TRUE or FALSE?

I say TRUE because the limit of g(x) is zero. In other words, I'm using the property above.

The author says FALSE. What am I missing here?

:?

 

[stapel]

In your first box, you've got a rule saying that, as long as g(x) is non-zero, then you can "split" the limit into to; that is, that the limit of [f(x)/g(x)] can be found by taking the limits of f(x) and g(x) separately.

In the second box, however, you can't use the rule from the first box, because g(x) does equal zero at the limit point of x = 5. So you can't split the thing up (which would create the division-by-zero problem you're thinking of).

Note that the author isn't saying that the limit in the second box does exist; he's only saying that it's false to say that the limit must necessarily not exist. It might. For instance:

. . . . .f(x) = (x - 5)(x + 3)

. . . . .g(x) = (x - 5)(x + 1)

. . . . .[f(x)/g(x)] = (x + 3)/(x + 1) for all x not equal to 5

So the limits of f(x) and g(x) as x goes to 5 are each zero, but the limit of the quotient is actually 4/3.

Hope that helps.

Eliz.

 

[DrToddMatthews]

Thank You Very Much

Note that the author isn't saying that the limit in the second box does exist; he's only saying that it's false to say that the limit must necessarily not exist.

Hope that helps.

Hello Elizabeth.

Your explanation and example help very much. As usual, I had tunnel vision. I thought "does not exist" could only mean "does exist".