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Given $f(x) = \frac{\sin x}{|x|^\alpha}$ with $\alpha >0$, Find $\lim\limits_{x \rightarrow \infty} f(x)$ if it exists. Justify using directly the following definition:

$\lim\limits_{x \rightarrow \infty} f(x)=L$ if $f$ is defined on an interval $(a, \infty)$ and for each $\epsilon >0$ there is a number $\beta$ such that: $$| f(x) - L| < \epsilon \quad\text{if}\quad x> \beta.$$

First, $$\lim\limits_{x \rightarrow \infty}\frac{\sin x}{|x|^\alpha}= \frac{\lim\limits_{x \rightarrow \infty}\sin x}{\lim\limits_{x \rightarrow \infty}|x|^\alpha} =0 $$

since the numerator oscillates from $-1$ to $1$ and denominator approaches $\infty$ with $\alpha >0.$

Second,

$$\left|\frac{\sin x}{|x|^\alpha} -0\right|< \epsilon$$ $$\left|\frac{\sin x}{|x|^\alpha}\right|< \epsilon.$$

We want to ensure that $\epsilon >\left|\frac{\sin x}{|x|^\alpha}\right|> 0.$

Case 1: If $x>0$, such that $x \in (0, k\pi)$ with $k \in \mathbb{Z}^{+}$, then $$\left|\frac{\sin x}{|x|^\alpha}\right| =\frac{\sin x}{x^\alpha}$$ it follows that: $$\left|\frac{\sin x}{|x|^\alpha}\right| = \frac{\sin x}{|x|^\alpha}<\epsilon \implies x>\frac{\sin x}{\epsilon}.$$

Case 2: If $x>0$ such that $x \in (k \pi, 2k\pi)$ then, $$\left|\frac{\sin x}{|x|^\alpha}\right| =\frac{-\sin x}{x^\alpha}< \epsilon \implies x> \frac{-\sin x}{\epsilon}.$$

Case 3: If $x<0$ such that $x \in (0, -k\pi)$ then, $$\left|\frac{\sin x}{|x|^\alpha}\right| =\left|\frac{\sin x}{-x^\alpha}\right|=\frac{\sin x}{x^\alpha}< \epsilon \implies x> \frac{\sin x}{\epsilon}.$$

Case 4: If $x<0$ such that $x \in (-k\pi, -2k\pi)$ then, $$\left|\frac{\sin x}{|x|^\alpha}\right| =\left|\frac{\sin x}{-x^\alpha}\right|=\frac{\sin x}{x^\alpha}< \epsilon \implies x> \frac{\sin x}{\epsilon}.$$

Am I going in the right direction? Any input is much appreciated.

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Whenever there is an $\epsilon, \delta$ or $\epsilon, N$ question, do not try to solve the inequation $|f(x) - L|<\epsilon$. Because we are not supposed to solve it (that is find all values of $x$ satisfying it) but rather we are supposed to ensure that the inequality $|f(x) - L |<\epsilon$ holds if the values of $x$ are constrained in a certain way.

The right approach is to do heavy simplification of the inequality $|f(x) -L|<\epsilon$ by using some inequality of the form $|f(x) - L|\leq g(x)$ where $g(x) $ is a very simple expression with perhaps bare minimum operations on $x$. And then try to ensure that $g(x) <\epsilon$. Sometimes finding an inequality like $|f(x) - L|\leq g(x) $ will require some obvious constraints on $x$ and we desire these constraints to be of the form $x>M$ where $M$ is some explicit positive number rather than just being a symbol.

In the current question we are so lucky to have the automatic inequality $$\left|\frac{\sin x} {x^{\alpha}} \right|\leq\frac{1}{x^{\alpha}}$$ The job is not done yet and we can do even more simplification by considering two cases $\alpha\geq 1$ and $0<\alpha<1$. If $\alpha\geq 1$ then $1/x^{\alpha}\leq 1/x$ and we can thus choose $g(x) =1/x$. Now we can ensure that $1/x<\epsilon$ if $x>1/\epsilon=N(\text{say})$. And we are done in case of $\alpha\geq 1$.

If $0<\alpha<1$ then there is a positive integer $k$ such that $1/k<\alpha$ and therefore $1/x^{\alpha}<1/x^{1/k}$ and then we can choose $g(x) =1/x^{1/k}$ and ensure $1/x^{1/k}<\epsilon$ by having $x>1/\epsilon^{k}=N(\text{say})$ and again we are done. It should be possible to find integer $k$ as any integer greater than $1/\alpha $ and it is better to keep an explicit value of $k$.

Doing such simplification is the crux of such problems and in general the expression for $N$ or $\delta$ in terms of $\epsilon$ is very simple due to this procedure. In fact you will see that the expression for $N$ or $\delta$ in terms of $\epsilon$ normally does not involve anything beyond field operations.


Your question uses the symbol $\beta$ in definition of limit which is pretty non-standard. I have used the symbol $N$ instead of $\beta$ and the use of symbol $N$ is more common.

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You may be thinking about this a bit too hard. For any $x > 0$, we have $\lvert \sin(x) \rvert \le 1$ and so $$\lvert f(x) \rvert = \left \lvert \frac{\sin(x)}{\lvert x \rvert^\alpha} \right \rvert \le \frac 1 {\lvert x \rvert^\alpha}.$$ Given $\epsilon > 0$, choose $\beta = 1/{\epsilon^{1/\alpha}}.$ Then $$\lvert x \rvert > \beta \implies \lvert x \rvert > 1/\epsilon^{1/\alpha} \implies \lvert x \rvert^\alpha > 1/\epsilon\implies \frac{1}{\lvert x \rvert^\alpha} < \epsilon. $$ But since $\lvert f(x) \rvert \le 1/\lvert x \rvert^\alpha$, this shows $$\lvert x \rvert > \beta \,\,\, \implies \,\,\, \lvert f(x) \rvert < \epsilon.$$

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We have for $x>0$,

$$|f (x)|\le \frac {1}{x^\alpha } $$

so if $$\frac {1}{x^\alpha} <\epsilon$$ then $$|f (x)|<\epsilon $$

or if

$$x> \frac {1}{\epsilon^{\frac {1}{\alpha}}}$$

then $$|f (x)|<\epsilon $$ thus we can take $$\beta=\frac {1}{\epsilon^{\frac {1}{\alpha}} }$$

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It seems you go generally in the right direction, but there are a few small mistakes.

First, $|\frac{\sin(x)}{|x|^\alpha}|$ can be zero. Secondly $\frac{\sin(x)}{|x|^\alpha} < \epsilon$ implies $|x|^\alpha > \frac{\sin(x)}{\epsilon}$, not necessarily $|x| > \frac{\sin(x)}{\epsilon}$ because $|x|^\alpha < |x|$ if $|x| < 1$. You can circumvent this by taking $\beta > 1$ or taking the $\alpha^{th}$ root.

Overall, it would be more efficient to note that $|\frac{\sin{x}}{|x|^\alpha}| < |\frac{1}{|x|^\alpha}|$, so $|\frac{\sin{x}}{|x|^\alpha}| < \epsilon$ is equivalent with $|x|^\alpha > \frac{1}{\epsilon}$ and then find a value of $\beta$ from there. You don't need to look at different cases this way. By choosing $\beta > 0$ at least, you can also leave out the absolute value signs.

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Fix $\varepsilon>0$. Then $$ |x|\ge \varepsilon^{-1/\alpha} \implies \left|\frac{\sin x}{x^\alpha}-0\right| \le \left|\frac{1}{x^\alpha}\right|<\varepsilon, $$ hence the limit is $0$.

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