Trouble with limits

Spec · #1 $Old$ November 2nd, 2007, 09:26 PM

I've encountered some difficult problems in my textbook that I can't solve. Any help would be appreciated. This is not homework—I'm just looking to advance my knowledge, that's all.

Question 1: (SOLVED, but I still don't understand the general method for solving these kinds of problems)
For each $\epsilon > 0$ , find a $\omega$ so that $x > \omega \implies \left| {\frac{\sqrt{x^2 + 1}}{x} - 1 }\right| < \epsilon$ is true. What does this show?

(By the way, how do I get the absolute characters to extend all the way in LaTeX?) SOLVED: Use \left| and \right|

Question 2: (SOLVED, but I still don't understand the general method for solving these kinds of problems)
Show that $x^3 \to 1$ when $x \to 1$ using the definition of limits.

This is what I've got so far:

As soon as we've chosen a $\epsilon > 0$ , then we'll be able to find a $\delta > 0$ so that
$|x^3 - 1| < \epsilon$ when $0 < |x - 1| < \delta$ .

Note:
$|x^3 - 1| = |x - 1| \cdot |x^2 + x + 1|$

Question 3: (SOLVED)
Is $f(x) = \lim_{n \to \infty} \frac{x^n}{1 + x^n}$ continuous for all $x \geq 0$ ?

Is $f(x) = \lim_{n \to \infty} \frac{x + x^2 + ... + x^n}{1 + x + ... + x^n}$ continuous for all $x \geq 0$ ?

kalagota · #2 $Old$ November 3rd, 2007, 12:09 AM

Quote:

Originally Posted by Spec $View Post$

Question 2:
Show that $x^3 \to 1$ when $x \to 1$ using the definition of limits.

This is what I've got so far:

As soon as we've chosen a $\epsilon > 0$ , then we'll be able to find a $\delta > 0$ so that
$|x^3 - 1| < \epsilon$ when $0 < |x - 1| < \delta$ .

Note:
$|x^3 - 1| = |x - 1| \cdot |x^2 + x + 1|$

if $|x| < 1 \, or \, (-1 < x < 1) \implies 0 < x^2 < 1$ so that

$-1 < x^2 + x < 2 \implies 0 < x^2 + x + 1 < 3$

$\implies |x^3 - 1| = |x - 1| \cdot |x^2 + x + 1| < 3\delta$
then choose $\delta = \frac{\epsilon}{3}$

topsquark · #3 $Old$ November 3rd, 2007, 05:12 AM

Quote:

Originally Posted by Spec $View Post$

Is $f(x) = \lim_{n \to \infty} \frac{x^n}{1 + x^n}$ continuous for all $x \geq 0$ ?

Consider x = 0. Now consider all other x. Can you define an epsilon and delta around x = 0?

-Dan

red_dog · #4 $Old$ November 3rd, 2007, 08:27 AM

If $x\in[0,1)$ then $\lim_{n\to\infty}x^n=0\Rightarrow f(x)=0$ .
If $x=1\Rightarrow f(1)=\frac{1}{2}$
If $\displaystyle x>1\Rightarrow\lim_{n\to\infty}\frac{x^n}{1+x^n}=\lim_{n\to\infty}\frac{x^n}{x^n\left[\left(\frac{1}{x}\right)^n+1\right]}=1$ .
So $\displaystyle f(x)=\left\{\begin{array}{cc}0, & x\in[0,1)\\\frac{1}{2}, & x=1\\1, & x>1\end{array}\right.$ , and f is discontinuous in x=1.

For the second one:
$f(x)=\left\{\begin{array}{cc}x, & x\in[0,1)\\\frac{n}{n+1}, & x=1\\1, & x>1\end{array}\right.$ and f is discontinuous in x=1.

Spec · #5 $Old$ November 3rd, 2007, 01:38 PM

Quote:

Originally Posted by red_dog $View Post$

If $x\in[0,1)$ then $\lim_{n\to\infty}x^n=0\Rightarrow f(x)=0$ .
If $x=1\Rightarrow f(1)=\frac{1}{2}$
If $\displaystyle x>1\Rightarrow\lim_{n\to\infty}\frac{x^n}{1+x^n}=\lim_{n\to\infty}\frac{x^n}{x^n\left[\left(\frac{1}{x}\right)^n+1\right]}=1$ .
So $\displaystyle f(x)=\left\{\begin{array}{cc}0, & x\in[0,1)\\\frac{1}{2}, & x=1\\1, & x>1\end{array}\right.$ , and f is discontinuous in x=1.

For the second one:
$f(x)=\left\{\begin{array}{cc}x, & x\in[0,1)\\\frac{n}{n+1}, & x=1\\1, & x>1\end{array}\right.$ and f is discontinuous in x=1.

The second one is supposed to be continuous for all $x \geq 0$ according to the answer in the textbook.

ThePerfectHacker · #6 $Old$ November 3rd, 2007, 05:41 PM

Quote:

Originally Posted by Spec $View Post$

I've encountered some difficult problems in my textbook that I can't solve. Any help would be appreciated. This is not homework—I'm just looking to advance my knowledge, that's all.

Question 1:
For each $\epsilon > 0$ , find a $\omega$ so that $x > \omega \implies |{\frac{\sqrt{x^2 + 1}}{x} - 1|} < \epsilon$ is true. What does this show?

This would show that,
$\lim_{x\to \infty} \frac{\sqrt{x^2+1}}{x} = 1$ .

Note that if $x>0$ then:
$\left| \frac{\sqrt{x^2+1}}{x} - 1 \right| = \left| \frac{\sqrt{x^2+1}-x}{x} \cdot \frac{\sqrt{x^2+1} + x}{\sqrt{x^2+1}+x} \right|$ $= \frac{1}{x(\sqrt{x^2+1}+x)} \leq \frac{1}{x(\sqrt{x^2}+x)} = \frac{1}{2x^2} \leq \frac{1}{x^2}$ .
So chose,
$C > \sqrt{\frac{1}{\epsilon}}$ .

ThePerfectHacker · #7 $Old$ November 3rd, 2007, 05:43 PM

Quote:

Originally Posted by Spec $View Post$

The second one is supposed to be continuous for all $x \geq 0$ according to the answer in the textbook.

The book is wrong.

Spec · #8 $Old$ November 3rd, 2007, 06:21 PM

Quote:

Originally Posted by ThePerfectHacker $View Post$

This would show that,
$\lim_{x\to \infty} \frac{\sqrt{x^2+1}}{x} = 1$ .

Note that if $x>0$ then:
$\left| \frac{\sqrt{x^2+1}}{x} - 1 \right| = \left| \frac{\sqrt{x^2+1}-x}{x} \cdot \frac{\sqrt{x^2+1} + x}{\sqrt{x^2+1}+x} \right|$ $= \frac{1}{x(\sqrt{x^2+1}+x)} \leq \frac{1}{x(\sqrt{x^2}+x)} = \frac{1}{2x^2} \leq \frac{1}{x^2}$ .
So chose,
$C > \sqrt{\frac{1}{\epsilon}}$ .

I have no idea what you did there. I'm with you right until this point (why did you choose to compare it to that?):

$\leq \frac{1}{x(\sqrt{x^2}+x)} = \frac{1}{2x^2} \leq \frac{1}{x^2}$ .
So chose,
$C > \sqrt{\frac{1}{\epsilon}}$ .

kalagota · #9 $Old$ November 3rd, 2007, 08:36 PM

because you are sure that the latter function is greater than the absolute thing, and it easier to compute, if that latter function is less than epsilon, then your absolute thing is less than epsilon..

red_dog · #10 $Old$ November 4th, 2007, 03:10 AM

Quote:

Originally Posted by Spec $View Post$

The second one is supposed to be continuous for all $x \geq 0$ according to the answer in the textbook.

Sorry, you're right. I forgot to take the limit in the case x=1.
$\displaystyle\lim_{n\to\infty}\frac{n}{n+1}=1$ .
So, $f(x)=\left\{\begin{array}{cc}x, & x\in[0,1)\\1, & x\geq 1\end{array}\right.$
and f is continuous for all $x\geq 0$

kalagota · #11 $Old$ November 4th, 2007, 05:47 AM

Quote:

Originally Posted by Spec $View Post$

...
Question 1: (SOLVED, but I still don't understand the general method for solving these kinds of problems)
For each $\epsilon > 0$ , find a $\omega$ so that $x > \omega \implies \left| {\frac{\sqrt{x^2 + 1}}{x} - 1 }\right| < \epsilon$ is true. What does this show?

(By the way, how do I get the absolute characters to extend all the way in LaTeX?) SOLVED: Use \left| and \right|

Question 2: (SOLVED, but I still don't understand the general method for solving these kinds of problems)
Show that $x^3 \to 1$ when $x \to 1$ using the definition of limits.

This is what I've got so far:

As soon as we've chosen a $\epsilon > 0$ , then we'll be able to find a $\delta > 0$ so that
$|x^3 - 1| < \epsilon$ when $0 < |x - 1| < \delta$ .

Note:
$|x^3 - 1| = |x - 1| \cdot |x^2 + x + 1|$

...

you'll just get used to it.. Ü

Spec · #12 $Old$ November 14th, 2007, 07:36 AM

I need help with another limit.

$\lim_{x \to \infty}\frac{x + ln(e^{2x} + x)}{\sqrt{x} + e^{1 + ln(x)}}$

This is what I've done so far.

$\lim_{x \to \infty}\frac{x + ln(e^{2x} + x)}{\sqrt{x} + e^{1 + ln(x)}} = \lim_{x \to \infty}\frac{\frac{x}{ln(e^{2x} + x)} + 1}{\frac{\sqrt{x}}{e^{1 + ln(x)}} + 1} \cdot \frac{ln(e^{2x} + x)}{e^{1 + ln(x)}}$

If this is correct, then the question becomes; what can I do with:

$\lim_{x \to \infty}\frac{ln(e^{2x} + x)}{e^{1 + ln(x)}}$

Also, I've solved this one, but my solution seems a bit strange. Anyone have a better solution?

$\lim_{x \to \infty}(1 + \frac{1}{5x})^{7x}$

I used a substitution:
$t = \frac{1}{x}$ , $t \to 0$ when $x \to \infty$

So we have:
$\lim_{t \to 0} (1 + \frac{t}{5})^{\frac{7}{t}} = \lim_{t \to 0} e^{ln(1 + \frac{t}{5})^{\frac{7}{t}}} = \lim_{t \to 0} e^{\frac{7}{5} \cdot ln(1 + \frac{t}{5})^{\frac{7}{t}^\frac{5}{7}}} = \lim_{t \to 0} e^{\frac{7}{5} \cdot ln(1 + \frac{t}{5})^{\frac{5}{t}}} = \lim_{t \to 0} e^{\frac{7}{5} \cdot ln(e)} = e^{\frac{7}{5}}$

topsquark · #13 $Old$ November 14th, 2007, 07:39 AM

Quote:

Originally Posted by Spec $View Post$

I need help with another limit.

$\lim_{x \to \infty}\frac{x + ln(e^{2x} + x)}{\sqrt{x} + e^{1 + ln(x)}}$

This is what I've done so far.

$\lim_{x \to \infty}\frac{x + ln(e^{2x} + x)}{\sqrt{x} + e^{1 + ln(x)}} = \lim_{x \to \infty}\frac{\frac{x}{ln(e^{2x} + x)} + 1}{\frac{\sqrt{x}}{e^{1 + ln(x)}} + 1} \cdot \frac{ln(e^{2x} + x)}{e^{1 + ln(x)}}$

If this is correct, then the question becomes; what can I do with:

$\lim_{x \to \infty}\frac{ln(e^{2x} + x)}{e^{1 + ln(x)}}$

Here's a possible route: Have you noticed that $e^{1 + ln(x)} = ex$ ? Then you can use ThePerfectHacker's favorite method: the squeeze theorem.

-Dan

Spec · #14 $Old$ November 14th, 2007, 11:54 AM

I'm not quite sure how to apply the squeeze theorem here. Are there any other solutions that doesn't require its use?

topsquark · #15 $Old$ November 14th, 2007, 12:25 PM

Quote:

Originally Posted by Spec $View Post$

$\lim_{x \to \infty}\frac{x + ln(e^{2x} + x)}{\sqrt{x} + e^{1 + ln(x)}}$

This is what I've done so far.

$\lim_{x \to \infty}\frac{x + ln(e^{2x} + x)}{\sqrt{x} + e^{1 + ln(x)}} = \lim_{x \to \infty}\frac{\frac{x}{ln(e^{2x} + x)} + 1}{\frac{\sqrt{x}}{e^{1 + ln(x)}} + 1} \cdot \frac{ln(e^{2x} + x)}{e^{1 + ln(x)}}$

If this is correct, then the question becomes; what can I do with:

$\lim_{x \to \infty}\frac{ln(e^{2x} + x)}{e^{1 + ln(x)}}$

Sorry, I don't know what I was thinking about for an upper bound on the limit.

What I would do is this:
$\lim_{x \to \infty}\frac{ln(e^{2x} + x)}{e^{1 + ln(x)}} = \frac{1}{e} \cdot \lim_{x \to \infty}\frac{ln(e^{2x} + x)}{x}$

This limit is of the form $\frac{\infty}{\infty}$ so we may use L'Hopital's rule here:
$\lim_{x \to \infty}\frac{ln(e^{2x} + x)}{e^{1 + ln(x)}} = \frac{1}{e} \cdot \lim_{x \to \infty}\frac{ln(e^{2x} + x)}{x} = \frac{1}{e} \cdot \lim_{x \to \infty} \frac{\frac{1}{e^{2x} + x} \cdot (2e^{2x} + 1)}{1}$ $= \frac{1}{e} \cdot \lim_{x \to \infty} \frac{2e^{2x} + 1}{e^{2x} + x}$

It should be clear that the value of this last limit is 2.

-Dan