Problem of the week #30 - October 22nd, 2012

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Jameson

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**Note - this for young students who might be studying pre-calculus or just beginning calculus and haven't been presented with the concept of upper/lower bounds and how they are useful in working with trig expressions.**

Consider $$\displaystyle \lim_{x \rightarrow 0} x^4 \cos \left( \frac{1}{x^2} \right)$$.

You can easily graph this and visually deduce that the answer is 0, which it is. Along the same lines you could make a table of (x,y) points and choose x's that get closer and closer to x=0 and again conclude that the limit is most likely 0. Although these two techniques might be useful on a test where you're stuck on a tricky limit, the answer can be found in a far more certain way.

So this week's question is prove that the above limit is 0.

Hint:
Start with the fact that $$\displaystyle -1 \le \cos(x) \le 1$$ with the goal of manipulating this somehow to where the middle term becomes the limit we want to find. Note that in the above, $\cos(x)$ cannot be smaller than -1 or larger than 1 for any value of x. This is true even if "x" is something more complicated. If you can't see how this applies in this problem, see hint 2.

Hint 2:
$$\displaystyle -1 \le \cos \left( \frac{1}{x^2} \right) \le 1$$ Now try manipulating this but make sure to do the same operation to all of the terms so it remains true, and remember that inequalities are not the same as equations and you must always keep positives and negatives in mind.

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Jameson

Staff member
Congratulations to the following members for their correct solutions:

1) Sudharaka

Solution (from Sudharaka):

$-1 \le \cos \left( \frac{1}{x^2} \right) \le 1$

Since $$x^4\geq 0$$ we have,

$\Rightarrow -x^4\le x^4 \cos \left( \frac{1}{x^2} \right) \le x^4$

When $$x\rightarrow 0$$ by the Squeeze Theorem (Squeeze theorem - Wikipedia, the free encyclopedia) we get,

$\lim_{x \rightarrow 0} x^4 \cos \left( \frac{1}{x^2} \right)=0$

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