If anyone can offer me some help...

**onmiverse** · April 30th 2009, 08:12 PM

I would really appreciate if anyone could lend me a hand (and a brain) to solve some of these questions
where something similar may appear in my upcoming final!

-Without using l'hopital's rule find the limit of tan(2x)/tan((pi)x) (sorry, I dont know how to type the pi sign)

-how can I prove that functions cannot have two different limits at the same point?

-Find the limit of sec(x sec^2x - tan^2x - 1) as X --> 1 (this question is particularly killing me)

-Without using derivatives show that the equation x^3 - 15x + 1 = 0 has three solutions in the interval ( -4, 4 )

Thankyou very much for reading this!!! I hope I don't bother anybody!

**matheagle** · April 30th 2009, 08:51 PM

Rewrite ${\tan(2x)\over \tan(\pi x)} = {\sin(2x) \cos(\pi x)\over \sin(\pi x) \cos(2x)}\sim {\sin(2x) \over \sin(\pi x)}$

SINCE, I am assuming x is heading towards zero and we know that cosine of zero is 1.

Also I plan to use (sin x)/x goes to 1 as x goes to 0.

Thus...

${\sin(2x) \over \sin(\pi x)} = \biggl({2\over \pi}\biggr) {(\sin(2x))/(2x) \over (\sin(\pi x))/(\pi x)}\to {2\over \pi}$ .

x^3 - 15x + 1 = 0 is a continuous function. So find some x's where the function is negative, positive...
By the intermediate value theorem (think, the price is right) the function must equal zero between those x values.

**redsoxfan325** · April 30th 2009, 08:51 PM

Originally Posted by onmiverse

I would really appreciate if anyone could lend me a hand (and a brain) to solve some of these questions
where something similar may appear in my upcoming final!

-Without using l'hopital's rule find the limit of $\frac{\tan(2x)}{\tan(\pi x)}$

-how can I prove that functions cannot have two different limits at the same point?

-Find the limit of $\sec(x \sec^2x - \tan^2x - 1)$ as $x\to1$ (this question is particularly killing me)

-Without using derivatives show that the equation $x^3 - 15x + 1 = 0$ has three solutions in the interval $( -4, 4 )$

Thankyou very much for reading this!!! I hope I don't bother anybody!

For the first question, is it as $x\to 0$ ?

------------------------------------

For the second question, let $\{x_n\}$ be a sequence that converges to $x$ and $x'$ . (We want to show that $x=x'$ .) Because of convergence, $\forall~\epsilon>0, \exists~ N$ such that $n>N \implies d(x,x_n)<\frac{\epsilon}{2}$ , and $\exists~ N'$ such that $n>N' \implies d(x_n,x')<\frac{\epsilon}{2}$ .

So if $n>\max\{N,N'\}$ , by the triangle inequality, $d(x,x')\leq d(x,x_n)+d(x_n,x')<\frac{\epsilon}{2}+\frac{\epsil on}{2} = \epsilon$ . Since $\epsilon$ was arbitrary, we can conclude that $x=x'$ .

----------------------------------

For the third question, you know this function is continuous at $x=1$ (because it is a composition of continuous functions), so you can just plug in $1$ for $x$ to get $\sec( 1\cdot\sec^2(1) - \tan^2(1) - 1)=\sec(0)=1$

------------------------------------

For the fourth question, $f(-4) = -3$ and $f(-3) = 19$ , so by the Intermediate Value Theorem $\exists~x_1\in(-4,-3)$ such that $f(x_1)=0$ .

Similarly, $f(1)=-13$ so the IMT says that $\exists~x_2\in(-3,1)$ such that $f(x_2)=0$ .

Similarly, $f(4)=5$ so the IMT says that $\exists~x_3\in(1,4)$ such that $f(x_3)=0$ .

Thus there are 3 zeroes in $(-4,4)$ .

--------------------------------

I hope I was some help.

**redsoxfan325** · April 30th 2009, 09:17 PM

Originally Posted by matheagle

Rewrite ${\tan(2x)\over \tan(\pi x)} = {\sin(2x) \cos(\pi x)\over \sin(\pi x) \cos(2x)}\sim {\sin(2x) \over \sin(\pi x)}$

SINCE, I am assuming x is heading towards zero and we know that cosine of zero is 1.

Also I plan to use (sin x)/x goes to 1 as x goes to 0.

Thus...

${\sin(2x) \over \sin(\pi x)} = \biggl({\pi\over 2}\biggr) {(\sin(2x))/(2x) \over (\sin(\pi x))/(\pi x)}\to {\pi\over 2}$ .

x^3 - 15x + 1 = 0 is a continuous function. So find some x's where the function is negative, positive...
By the intermediate value theorem (think price is right) the function must equal zero between those x values.

I think it should be $\left(\frac{2}{\pi}\right)\cdot\frac{\frac{\sin(2x )}{2x}}{\frac{\sin(\pi x)}{\pi x}}\to\frac{2}{\pi}$

**onmiverse** · April 30th 2009, 09:48 PM

Originally Posted by redsoxfan325

For the first question, is it as $x\to 0$ ?

------------------------------------

For the second question, let $\{x_n\}$ be a sequence that converges to $x$ and $x'$ . (We want to show that $x=x'$ .) Because of convergence, $\forall~\epsilon>0, \exists~ N$ such that $n>N \implies d(x,x_n)<\frac{\epsilon}{2}$ , and $\exists~ N'$ such that $n>N' \implies d(x_n,x')<\frac{\epsilon}{2}$ .

So if $n>\max\{N,N'\}$ , by the triangle inequality, $d(x,x')\leq d(x,x_n)+d(x_n,x')<\frac{\epsilon}{2}+\frac{\epsil on}{2} = \epsilon$ . Since $\epsilon$ was arbitrary, we can conclude that $x=x'$ .

----------------------------------

For the third question, you know this function is continuous at $x=1$ (because it is a composition of continuous functions), so you can just plug in $1$ for $x$ to get $\sec( 1\cdot\sec^2(1) - \tan^2(1) - 1)=\sec(0)=1$

------------------------------------

For the fourth question, $f(-4) = -3$ and $f(-3) = 19$ , so by the Intermediate Value Theorem $\exists~x_1\in(-4,-3)$ such that $f(x_1)=0$ .

Similarly, $f(1)=-13$ so the IMT says that $\exists~x_2\in(-3,1)$ such that $f(x_2)=0$ .

Similarly, $f(4)=5$ so the IMT says that $\exists~x_3\in(1,4)$ such that $f(x_3)=0$ .

Thus there are 3 zeroes in $(-4,4)$ .

--------------------------------

I hope I was some help.

Thank you really all of you guys who replied... but I don't get the second answer...if you could expand upon that I would greatly appreciated it...

**matheagle** · April 30th 2009, 09:48 PM

Yup, I had it up-side down.

${\sin(ax) \over \sin(b x)} \to {a\over b}$ as x heads towards zero.

**matheagle** · April 30th 2009, 09:52 PM

Originally Posted by onmiverse

Thank you really all of you guys who replied... but I don't get the second answer...if you could expand upon that I would greatly appreciated it...

Draw f(x) via just a few points.
Find some where f is positive and some in your interval where f is negative.
Since f is continuous, it must cross zero between those x's.

**redsoxfan325** · April 30th 2009, 10:03 PM

I'm not sure what calculus class you're in, but the proof I gave you was one I learned in Analysis, which is essentially a course entirely devoted to proving all the theorems that make up calculus.

Some notation: $\epsilon$ is generally used to represent an arbitrary distance, usually one that is very small. $d(p,q)$ is the distance (a real number) between $p$ and $q$ .

The idea behind this proof is that for large enough $n$ , the distance between any given point in the sequence and the point to which the sequence converges becomes very small. So if the sequence converges to $x$ and $x'$ , then for large enough $n$ , $x_n$ gets very close to $x$ and $x'$ . So if for large $n$ , $d(x_n,x)\approx0$ and $d(x_n,x')\approx0$ , then it makes sense that $d(x,x')\approx0$ as well. If two points have zero distance between them, then they must be the same point. Thus $x=x'$ .

So intuitively, that's why the proof works.

**onmiverse** · April 30th 2009, 10:11 PM

I get it now! Once it was explained with words I could easilly understand, and again, I am ever thankful for your generosity...

**matheagle** · April 30th 2009, 10:15 PM

My guess is that this is a calc 1 or 2 student and he/she had no idea what an epsilon is, besides a frat symbol.

Originally Posted by redsoxfan325

I'm not sure what calculus class you're in, but the proof I gave you was one I learned in Analysis, which is essentially a course entirely devoted to proving all the theorems that make up calculus.

Some notation: $\epsilon$ is generally used to represent an arbitrary distance, usually one that is very small. $d(p,q)$ is the distance (a real number) between $p$ and $q$ .

The idea behind this proof is that for large enough $n$ , the distance between any given point in the sequence and the point to which the sequence converges becomes very small. So if the sequence converges to $x$ and $x'$ , then for large enough $n$ , $x_n$ gets very close to $x$ and $x'$ . So if for large $n$ , $d(x_n,x)\approx0$ and $d(x_n,x')\approx0$ , then it makes sense that $d(x,x')\approx0$ as well. If two points have zero distance between them, then they must be the same point. Thus $x=x'$ .

So intuitively, that's why the proof works.

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thank you very much, but I still don't understand...

thank you once again

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