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I have a function $g(x)=x^4 \cos(2/x)$.

I have to use Squeeze Theorem to show $\lim\limits_{x\to0}g(x)=0$.

Usually all the questions I have done up till now on Squeeze theorem have provided me with 3 functions, 1 function is less than another. 1 function is the highest of them all. and 1 functino is in between the 2 functions. From there I have been able to prove the limit with squeeze theorem but in here I have no other function so what do I do here?

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    $\begingroup$ Your job is to find the upper and lower functions. Noting that $\cos(x)$ is bounded between two values should help you get started with finding the appropriate functions. $\endgroup$ – JavaMan Nov 30 '11 at 17:12
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You can use the fact that $\lim\limits_{x\to a}f(x) = 0$ iff $\lim\limits_{x\to a}|f(x)| = 0$, so in your case: $0\leq|f(x)| = x^4|\cos{\frac2x}|\leq x^4$ Since both $0$ and $x^4$ goes to zero with $x\to 0$ you obtain what you want. In the case you don't want to deal with the absolute value - just note that $-x^4\leq x^4\cos\frac2x\leq x^4$ since $-1\leq\cos \frac2x\leq1$ for all $x\neq 0$ as JavaMan has pointed in his comment.

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Since $x^4$ is never negative, you can say that since $-1\le\cos(2/x)\le1$, it follows that $-x^4\le x^4\cos(2/x)\le x^4$. There you have your three functions.

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