# infinite/finite limits

• Mar 9th 2009, 02:50 PM
buttonbear
infinite/finite limits
which of these limits is finite?

A) lim 1.001^n
n->infinity

B) lim ln (1/1+t)
t->infinity

C) lim (e^x)/x
x->infinity

D) lim (1 + x/2)^(1/x)
x->0

i'm sorry that they're not in the nicest format
i figured that the first two were just infinity, but i'm not sure..
• Mar 9th 2009, 02:57 PM
Prove It
Quote:

Originally Posted by buttonbear
which of these limits is finite?

A) lim 1.001^n
n->infinity

B) lim ln (1/1+t)
t->infinity

C) lim (e^x)/x
x->infinity

D) lim (1 + x/2)^(1/x)
x->0

i'm sorry that they're not in the nicest format
i figured that the first two were just infinity, but i'm not sure..

The first three are infinite.

A) Any number $> 1 \to \infty$ as it gets exponentiated.

B) The inside of the brackets $\to \frac{1}{\infty} \to 0$.

$\log{x} \to -\infty$ as $x \to 0$.

C) The top and bottom both tend to $\infty$. This is indeterminate, but we can use L'Hospital.

$\lim_{x \to \infty}\frac{e^x}{x} = \lim_{x \to \infty}\frac{\frac{d}{dx}e^x}{\frac{d}{dx}x} = \lim_{x \to \infty}\frac{e^x}{1} \to \infty$.

• Mar 9th 2009, 03:26 PM
Jester
If the last one is

$\lim_{x \to 0} \left( 1 + \frac{x}{2} \right)^{\frac{1}{x}}$

then

$\lim_{x \to 0} e^{ \ln \left( 1 + \frac{x}{2} \right)^{\frac{1}{x}}} = \lim_{x \to 0} e^{ \frac{1}{x}\ln \left( 1 + \frac{x}{2} \right)}$

which comes down to finding

$\lim_{x \to 0} \frac{1}{x}\ln \left( 1 + \frac{x}{2} \right) = \frac{1}{2}$.

using L'Hopital's rule, so

$\lim_{x \to 0} \left( 1 + \frac{x}{2} \right)^{\frac{1}{x}} =
\sqrt{e}$
• Mar 9th 2009, 03:36 PM
buttonbear
i'm a little confused by what happens in the last two lines of the last limit- would you mind explaining?(Thinking)
• Mar 10th 2009, 05:04 AM
Jester
Quote:

Originally Posted by buttonbear
i'm a little confused by what happens in the last two lines of the last limit- would you mind explaining?(Thinking)

Sure. Using L'Hoptials rule

$\lim_{x \to 0} \frac{1}{x}\ln \left( 1 + \frac{x}{2} \right) = \lim_{x \to 0} \frac{1}{1 + \frac{x}{2} } \frac{1}{2} = \frac{1}{2}$

so

$\lim_{x \to 0} e^{ \ln \left( 1 + \frac{x}{2} \right)^{\frac{1}{x}}} = \lim_{x \to 0} e^{ \frac{1}{x}\ln \left( 1 + \frac{x}{2} \right) }= e^{ \lim_{x \to 0} \frac{1}{x}\ln \left( 1 + \frac{x}{2} \right) }= e^{1/2}$
• Mar 10th 2009, 06:19 AM
buttonbear
$\lim_{x \to 0} \frac{1}{x}\ln \left( 1 + \frac{x}{2} \right) = \frac{1}{2}$

so you used l'hopital's rule here?
i understand everything else..i'm just not getting the 1/2
• Mar 10th 2009, 06:45 AM
skeeter
Quote:

Originally Posted by buttonbear
$\lim_{x \to 0} \frac{1}{x}\ln \left( 1 + \frac{x}{2} \right) = \frac{1}{2}$

so you used l'hopital's rule here?
i understand everything else..i'm just not getting the 1/2

$\lim_{x \to 0} \frac{\ln \left( 1 + \frac{x}{2} \right)}{x}$

L'Hopital ... $\lim_{x \to 0} \frac{f'(x)}{g'(x)}$ ...

$\lim_{x \to 0} \frac{\frac{\frac{1}{2}}{1 + \frac{x}{2}}}{1} = \frac{1}{2}$
• Mar 10th 2009, 06:50 AM
buttonbear
i guess sometimes i just need things spelled out for me. thanks so much!
• Mar 10th 2009, 06:51 AM
Jester
Quote:

Originally Posted by buttonbear
$\lim_{x \to 0} \frac{1}{x}\ln \left( 1 + \frac{x}{2} \right) = \frac{1}{2}$

so you used l'hopital's rule here?
i understand everything else..i'm just not getting the 1/2

If $f = \ln \left(1 + \frac{x}{2} \right)\, \text{and}\; g = x$ then $f' = \frac{1}{1 + \frac{x}{2}} \cdot \frac{1}{2}\; \text{and}\; g' = 1$ (the first by the chain rule). so

$\lim_{x \to 0} \frac{f'}{g'} = \lim_{x \to 0} \frac{1}{1 + \frac{x}{2}} \cdot \frac{1}{2} = \frac{1}{2}$

This is where the 1/2 comes from. Better?
• Mar 10th 2009, 10:36 AM
buttonbear
much better- thanks so much!