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$$\lim_{x\to \infty}\frac{5x^2-4}{x-x^2} = \frac{\infty}{\infty}$$

I know I will have to use L'Hôpital's rule to solve. However, I'm confused as to how the infinity sign is calculated.

At first I thought the infinity sign to be just a large number. The only difference being whether it was a large positive number or a large negative number.

This number multiplied by a constant either increased or decreased the constant without bounds.

However, how should I calculate $5x^2-4$?

$5 \cdot \infty^2-4 = $ enter image description here

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  • $\begingroup$ Hint: Divide the numerator and denominator by $x^{2}$ and then evaluate the limit. $\endgroup$ – mattos Dec 6 '15 at 14:49
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    $\begingroup$ With the abbreviation $\infty/\infty$ one means that the limits of the numerator and the denominator are $\pm\infty$. $\endgroup$ – egreg Dec 6 '15 at 14:51
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    $\begingroup$ Hint: $\lim_{x\to\infty}\frac{5x^2-4}{x-x^2}$ does not equal the undefined expression $\frac\infty\infty$, rather this is a limit of indeterminate type $\frac\infty\infty$ $\endgroup$ – Hagen von Eitzen Dec 6 '15 at 14:52
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    $\begingroup$ @Sunny What's correct is $\lim_{x\to\infty}(5x^2-4)=\infty$. You don't multiply $5$ times infinity. $\endgroup$ – egreg Dec 6 '15 at 14:55
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    $\begingroup$ Yes, I guess the point is L'Hopital works for indeterminate form $(+\infty)/(-\infty)$ but it may just be written $\infty/\infty$, especially by guys who know complex analysis. $\endgroup$ – GEdgar Dec 6 '15 at 15:19
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The key thing is that infinity isn't a number in the traditional sense, so we need to carefully build up our intuition and how we think about infinity.

When we say $x$ tends to infinity, rigorously speaking, we mean that it gets arbitrarily large. For example, $x^2 \rightarrow \infty$ as $x \rightarrow \infty$ since, as $x$ gets arbitrarily large, $x^2$ gets arbitrarily large.

We can think of this graphically. Consider the graph $y=x^2$. As $x \rightarrow \infty$, the curve goes off to $\infty$ as well.

Similarly, $5x^2 - 4 \rightarrow \infty$ as $x \rightarrow \infty$, since $5x^2 - 4$ gets arbitrarily large as $x$ does (we see this graphically too).

The key thing to take away from this is that multiplying $x^2$ by $5$ didn't change the fact it's going to $\infty$. Neither did subtracting $4$. In fact, we could multiply, divide, add, or subtract any fixed number we want, but the expression is still going to go off to $\infty$ eventually.

However, if we're adding, subtracting, multiplying or dividing by things that aren't fixed, then it isn't obvious what's happening as $x \rightarrow \infty$. For instance, individually, $x$ and $-x^2$ both go off to $\pm \infty$ individually, but what happens to $x-x^2$?

In general, anything can happen when we have sums and quotients of things that go to $\infty$ separately. We call these 'indeterminate'. For indeterminate forms, we need to use other tricks and theorems to determine what's going on.

For example, for $x-x^2$, graphically we see that it goes to $-\infty$ as $x \rightarrow \infty$. This is because $x^2$ gets bigger much faster than $x$.

For the quotient in your question, there are a few tools at our disposal. Dr. Graubner's solution shows one such trick, and L'Hôpital is another.

Hope this helped with how to think about these things. I'd also advise going back through your notes (or the internet) and revising the rigorous definitions regarding what it means to tend to infinity.

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rewrite your term in the form $$\frac{5-\frac{4}{x^2}}{\frac{1}{x}-1}$$

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