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# Calculus Done Right

The mathematics of the continuous, through intuition not memorization.

We begin our study of limits by looking at the expression:

\[\frac{\sin(x)}{x}\]

This gives a well-defined number for every value of \(x,\) except \(x=0.\) So what can we say about the behavior of the expression at 0? We can't plug in 0 directly, but we can plug in very small, non-zero values of \(x.\) And we can plug in smaller and smaller values, and see if there is a pattern in the value of the expression as \(x\) *approaches 0*.

Investigate what happens to \(\frac{\sin(x)}{x}\) as \(x\) gets small. Try, for example, \(x=1,\) \(x=0.1,\) and \(x=0.01.\) What seems to be happening to \(\frac{\sin(x)}{x}\)?

Use a calculator. There are plenty of online calculators, like this one.

**Note:** Make sure you are evaluating \(\sin\) in *radian mode*.

We saw that as \(x\) gets smaller and smaller, the expression \(\frac{\sin(x)}{x}\) gets closer and closer to 1. We'll have a lot more to say about the details, but this is the basic idea behind saying "the limit as \(x\) approaches 0 of \(\frac{\sin(x)}{x}\) is 1." In symbols:

\[\lim_{x \to 0}\frac{\sin(x)}{x} = 1\]

Note that as \(x\) gets small, both the numerator and denominator get close to 0. This limit is an example of the \(\frac{0}{0}\) indeterminate form that we saw in the first Chapter. We’ll be encountering this form a lot.

You should have found that the expression isn't getting closer and closer to any particular number as \(x\) approaches 0. We say that \(\lim\limits_{x \to 0} \sin\left(\frac{1}{x}\right)\) does not exist.

Which of these arguments explains this behavior?

**A)** As \(x\) gets close to 0, \(\sin(x)\) also gets close to 0, and when we take the reciprocal of \(\sin(x)\), we get something that gets close to infinity, so it doesn't get closer and closer to any particular number.

**B)** As \(x\) gets close to 0, \(\sin(x)\) oscillates between 0 and 1, and so does the reciprocal, so it never gets close to any particular number.

**C)** When you actually plug in \(x=0,\) you get \(\sin\left(\frac{1}{0}\right),\) which is undefined, so the limit does not exist.

**D)** As \(x\) gets close to 0, \(\frac{1}{x}\) gets bigger and bigger, so we're putting a bigger and bigger number into \(\sin,\) and since \(\sin\) oscillates forever between -1 and 1 as its input gets large, the final result doesn't get closer to any particular number.

Let's investigate one final example: \(\lim\limits_{x \to 0} \; x \sin\left(\frac{1}{x}\right).\) What do you think happens to \(x \sin\left(\frac{1}{x}\right)\) as \(x\) gets small?

**A)** I think it doesn't approach any particular number, because, as we saw in the last question, the \(\sin\left(\frac{1}{x}\right)\) term oscillates between -1 and 1 as \(x\) gets small.

**B)** I think it gets closer and closer to 0, because even though \(\sin\left(\frac{1}{x}\right)\) is doing crazy things as \(x\) gets small, we're also multiplying by \(x,\) and that's going to 0.

**C)** I think it gets closer and closer to 1, but I'm not sure why.

Try it on your calculator!

Let's end this quiz by getting some graphical insight. The graph of \(y = \sin\left(\frac{1}{x}\right)\) looks like:

This is a strange graph, but we can see that as \(x\) gets small, the corresponding \(y\) values are alternating between \(-1\) and \(1\), faster and faster. So \(\displaystyle\lim_{x \to 0} \sin\left(\frac{1}{x}\right)\) does not exist.

But the graph of \(y = x \sin\left(\frac{1}{x}\right)\) looks like:

You can think of the \(x\) factor as an amplitude that's changing. Instead of oscillating between 0 and 1, this expression oscillates between the lines \(y=x\) and \(y= -x.\) And you can see that as \(x\) gets small, the corresponding \(y\) values get small too. So \(\displaystyle\lim_{x \to 0} x \sin\left(\frac{1}{x}\right) = 0.\)

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