I know that the limit laws state that the limit of a sum, difference, product, and quotient (if the limit of the denominator is not 0) is the [given operator] of the limits. However, I can't seem to prove or find whether this applies with limits that approach infinity.

Could someone verify whether I can apply the limit laws with limits that approach infinity? Thanks.

  • 1
    $\begingroup$ Um... I think that 0 times infinity is undefined. This is sort of like the exception to the quotient rule, I think. If the end result is defined, would the laws still sometimes fail? $\endgroup$ – Jed Dec 9 '15 at 4:11
  • 1
    $\begingroup$ Limit laws only apply when all of the limits involved exist (and the example illustrates why it can fail when they don't). $\endgroup$ – TokenToucan Dec 9 '15 at 4:11

I have yet to come across a calculus textbook that explains this concept, which just goes to show how bad calculus textbooks are.

You can use the limit laws with infinite limits so long as you don't end up with an "indeterminate form". Some common indeterminate forms that you will run across are ∞ - ∞, ∞ ∙ 0, and ∞/∞.

For example, lets say that your function is f(x) = x + x. The limit of f(x) as x→∞ can be evaluated as ∞ + ∞, which equals ∞. But, if the function was instead f(x) = x - x, then you would end up with ∞ - ∞, which is indeterminate, so the limit law doesn't apply.

Note that there is a difference between the idea of an infinite limit and a limit that approaches infinity (usually this is called a "limit at infinity"). A limit at infinity means x→∞. An infinite limit is when a limit evaluates to ∞. An example of an infinite limit is lim x→0 of ln(x), which evaluates to -∞. An example of a limit at infinity is lim x→∞ of (3x^2 + 1)/(x^2 - 7), which evaluates to 3. You can also have an infinite limit at infinity, which is what I did in the above paragraph.

For limits at infinity, all of the limit laws are the same as the regular limit laws so long as you don't have infinite limits.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.