Laters' Maths Help Thread (1 Viewer)

laters · Mar 19, 2016

Hey guys,

I don't come on here often but I've seen similar threads so I've decided to make one too

For your reference I am doing MATH1151 this sem.

1) Find $\lim_{x\rightarrow\infty} \sin\frac{1}{x}$ . I already know it's 0 but they want a solution using the pinching theorem.

InteGrand · Mar 19, 2016

laters said:
Hey guys,

I don't come on here often but I've seen similar threads so I've decided to make one too For your reference I am doing MATH1151 this sem.

1) Find $\lim_{x\rightarrow\infty} \sin\frac{1}{x}$ . I already know it's 0 but they want a solution using the pinching theorem.

$\noindent It's just equivalent to $\lim _{\theta \to 0^{+}} \sin \theta$, which is 0. If you really wanted to use the pinching theorem, you could use the fact that for $\theta \in \left[0,\frac{\pi}{2}\right]$ say, we have $0 \leq \sin \theta \leq \theta$. Then by the pinching theorem (aka squeeze law, and a whole host of other names), as $\theta \to 0^{+}$, $\sin \theta \to 0$.$

leehuan · Mar 19, 2016

InteGrand said:
$\noindent It's just equivalent to $\lim _{\theta \to 0^{+}} \sin \theta$, which is 0. If you really wanted to use the pinching theorem, you could use the fact that for $\theta \in \left[0,\frac{\pi}{2}\right]$ say, we have $0 \leq \sin \theta \leq \theta$. Then by the pinching theorem (aka squeeze law, and a whole host of other names), as $\theta \to 0^{+}$, $\sin \theta \to 0$.$

Is there an explicit way to evaluate this withOUT the (mental) substitution of u=1/x and only by the squeeze theorem?

InteGrand · Mar 19, 2016

leehuan said:
Is there an explicit way to evaluate this withOUT the (mental) substitution of u=1/x and only by the squeeze theorem?

$$\noindent Yes. For example, just note that $0\leq \sin \frac{1}{x}\leq \frac{1}{x}$ for all $x\in \left [\frac{2}{\pi},\infty \right)$ $\Big{(}$this is seen by essentially applying the map $x\mapsto \frac{1}{x}$ to our earlier interval and method$\Big{)}$, and then use the squeeze law as $x\to \infty$.$

leehuan · Mar 19, 2016

Oh...duh

laters · Mar 21, 2016

2) Suppose that for any K>2 the solution to f(x) > K is $x\epsilon(1,\frac{K}{K-2})[\tex] What is the behaviour of f(x) as x approaches 1+? Carefully explain your answer This is mainly a problem with my working. So intuitively I know that the limit must be infinity. the reason being that if K approaches infinity then K/(K-2) must approach 1+. And so for x between 1 and [B]"1+"[/B], [B]f(x)>infinity[/B]. and so the limit must be infinity. I'm just a little worried about the statements 1<x<1+ and f(x)>infinity. Obviously people would know what I mean but I would like an explanation that avoids saying these things. Perhaps using the delta-epsilon definition? Although working it into this example may be tricky idk$

InteGrand · Mar 21, 2016

laters said:
2) Suppose that for any K>2 the solution to f(x) > K is $x\epsilon(1,\frac{K}{K-2})[\tex] What is the behaviour of f(x) as x approaches 1+? Carefully explain your answer This is mainly a problem with my working. So intuitively I know that the limit must be infinity. the reason being that if K approaches infinity then K/(K-2) must approach 1+. And so for x between 1 and [B]"1+"[/B], [B]f(x)>infinity[/B]. and so the limit must be infinity. I'm just a little worried about the statements 1<x<1+ and f(x)>infinity. Obviously people would know what I mean but I would like an explanation that avoids saying these things. Perhaps using the delta-epsilon definition? Although working it into this example may be tricky idk[/QUOTE] [tex]$\noindent The answer is that $f(x) \to \infty$ as $x \to 1^{+}$ This follows from the epsilon-delta definition of ``$f(x) \to \infty$ as $x\to a^{+}$'', where $a\in \mathbb{R}$. For your reference, the definition is that for any $K>0$ there exists a $\delta >0$ such that whenever $a<x<a+\delta$, we have $f(x) > K$. In your particular example, this is satisfied for any $K>2$ (and hence by transitivity of inequality, for any $K>0$), by taking $\delta = \frac{2}{K-2}$, for $K>2$. This is because we are given that whenever $x \in \left(1, \frac{K}{K-2}\right)$ ($K>2$), we have $f(x) >K$. $\Big{(}$Note that $\frac{K}{K-2}=1+\frac{2}{K-2}\equiv 1 + \delta$.$\Big{)}$$$

laters · Apr 1, 2016

Let f be continuous in $\mathbb{R}$ with $\lim_{x\rightarrow\infty}f(x)=\lim_{x\rightarrow-\infty}f(x)=0$

Showthat if there is a number $\xi$ such that $f(\xi)>0$ then f attains a maximum value in the reals.
[Note the max min theorem applies to finite closed intervals [a,b] only]

Graphically I understand what they're saying... but I am struggling to write a more concrete proof.

InteGrand · Apr 1, 2016

laters said:
Let f be continuous in $\mathbb{R}$ with $\lim_{x\rightarrow\infty}f(x)=\lim_{x\rightarrow-\infty}f(x)=0$

Showthat if there is a number $\xi$ such that $f(\xi)>0$ then f attains a maximum value in the reals.
[Note the max min theorem applies to finite closed intervals [a,b] only]

Graphically I understand what they're saying... but I am struggling to write a more concrete proof.

If we can assume the extreme value theorem, we can do it as follows.
Let $f(\xi)=\Xi >0$ .

By definition of limits to +/- infinity, there exist real numbers $M>N$ such that $|f(x)-0|<\Xi$ whenever $x>M$ and whenever $x<N$ . So $\xi \in [N,M]$ (because outside this interval, $f(x)\neq \Xi$ ), and:

(1) $f(x)<\Xi \text{ }\forall x \in (-\infty, N)$

(2) $f(x)<\Xi \text{ }\forall x\in(M,\infty)$ .

By the extreme value theorem (since f is continuous), $f$ attains a maximum value $K$ on $[N,M]$ .

Since $f(x)=\Xi >0$ when $x= \xi \in [N,M]$ , this maximum $K$ satisfies $K\geq \Xi$ . (3)

(1), (2) and (3) imply that f attains a maximum value on $\mathbb{R}$ (doing so in the interval [N, M]), namely $f_\text{max.}=K$ .

Trebla · Apr 3, 2016

This thread will be closed shortly.

After careful consideration, it was decided that user specific threads will no longer be allowed (whether they be one user helping or one user asking for help). We have noticed that the first few threads have spawned multiple others, which is not desirable in the spirit of being an open community. The last thing we want is the Maths forums being full of "User X Maths Help" threads.

You are encouraged to post separate questions in separate threads, though keep in mind a lot of the questions that I have seen so far have been answered before so I suggest to actually do a search first to see if it hasn't been answered already in past threads (which is easier to do when there is a specific thread for a specific question or topic rather than a general user specific thread).

Laters' Maths Help Thread (1 Viewer)

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