# hard limit.

• February 17th 2011, 06:08 AM
Moklin
hard limit.
limf(x)
x--+oo

f(x)=(5^x)/(1-lnx)

i can´t solve this. i believe the answer is +00 but i would like to see the resolution step by step.
Sorry if i have posted in the wrong section.
• February 17th 2011, 06:19 AM
yeKciM
Quote:

Originally Posted by Moklin
limf(x)
x--+oo

f(x)=(5^x)/(1-lnx)

i can´t solve this. i believe the answer is +00 but i would like to see the resolution step by step.
Sorry if i have posted in the wrong section.

it's $-\infty$

use L'Hospitals rule .... and show what you get :D
• February 17th 2011, 06:26 AM
Moklin
Yeah. It´s -oo thanks, but we are not supposed to use L´Hospitals rule to solve it. :/
• February 17th 2011, 02:38 PM
sinx
This is how I was taught to do it without L'Hospitals rule

$\lim_{x \to +\infty} f(x) = \frac{(5^x)}{(1-lnx)} = - \infty$

The trick that I used is to observe all the terms as they approach a number that is getting infinitely larger. Since it's specified we only want it as it approaches positive infinity we don't need to observe it as it approaches a number that gets infinitely smaller, since the limit from the left and the right as it approaches positive infinity will be infinitely large.

So first I look at the top and see that 5 is raised to X. What happens to 5 as it's exponent gets infinitely larger? The result is infinitely larger, so I just substituted it with a + sign to remind myself it will stay positive. It now looks like:

$\lim_{x \to +\infty} f(x) = {\frac{+}{(1-lnx)}$

Then I observe the bottom. The limit of a constant is the constant, so we know as X gets infinitely larger 1 will stay 1.

The interesting behavior is at $ln(x)$. As x gets infinitely larger, it stays positive. You can test this by grabbing any calculator and taking the natural log of any extremely large positive number.

What is 1 subtract a large positive number? A negative of course, which gives me (using my symbology) -

$\frac{+}{-}$

which comes out to be a negative. This tells me that the limit will approach negative infinity due to the behavior in the denominator.

This trick works for simple stuff like this. Even if you won't learn L'Hospital's rule in your class until second semester calc, you should pick it up if not just to check your work.
• February 17th 2011, 02:51 PM
topsquark
Quote:

Originally Posted by sinx
This is how I was taught to do it without L'Hospitals rule

$\lim_{x \to +\infty} f(x) = \frac{(5^x)}{(1-lnx)} = - \infty$

The trick that I used is to observe all the terms as they approach a number that is getting infinitely larger. Since it's specified we only want it as it approaches positive infinity we don't need to observe it as it approaches a number that gets infinitely smaller, since the limit from the left and the right as it approaches positive infinity will be infinitely large.

So first I look at the top and see that 5 is raised to X. What happens to 5 as it's exponent gets infinitely larger? The result is infinitely larger, so I just substituted it with a + sign to remind myself it will stay positive. It now looks like:

$\lim_{x \to +\infty} f(x) = {\frac{+}{(1-lnx)}$

Then I observe the bottom. The limit of a constant is the constant, so we know as X gets infinitely larger 1 will stay 1.

The interesting behavior is at $ln(x)$. As x gets infinitely larger, it stays positive. You can test this by grabbing any calculator and taking the natural log of any extremely large positive number.

What is 1 subtract a large positive number? A negative of course, which gives me (using my symbology) -

$\frac{+}{-}$

which comes out to be a negative. This tells me that the limit will approach negative infinity due to the behavior in the denominator.

This trick works for simple stuff like this. Even if you won't learn L'Hospital's rule in your class until second semester calc, you should pick it up if not just to check your work.

This is good as far as it goes, but how to you handle which (if either) of the infinities are bigger than the other? In other words, you will have to prove that $\displaystyle \lim_{x \to \infty} 5^x$ is a "larger" infinity than $\displaystyle \lim_{x \to \infty} ln(x)$. You have not done so.

-Dan
• February 17th 2011, 02:58 PM
sinx
Quote:

Originally Posted by topsquark
This is good as far as it goes, but how to you handle which (if either) of the infinities are bigger than the other? In other words, you will have to prove that $\displaystyle \lim_{x \to \infty} 5^x$ is a "larger" infinity than $\displaystyle \lim_{x \to \infty} ln(x)$. You have not done so.

-Dan

I suppose the trick in it is that regardless of which infinity is larger, you're still subtracting something infinitely large from 1 resulting in a negative, and the number in the numerator will remain infinitely positive as nothing is acting on it but X. Even though it doesn't stand for proof, you could easily say that

$\frac{5,000,000}{1-5,999,999}$

will be some negative value simply because 1 subtract any large number will be negative. However, I still talk to the professor that taught me this trick so I'll ask him to prove it for me next week and see what happens. I'll post the result. Good job pointing it out, I didn't even consider proofing it.
• February 17th 2011, 03:14 PM
Quote:

Originally Posted by Moklin
limf(x)
x--+oo

f(x)=(5^x)/(1-lnx)

i can´t solve this. i believe the answer is +00 but i would like to see the resolution step by step.
Sorry if i have posted in the wrong section.

You could use the fact that

$e^x>lnx\Rightarrow\ e^x>lnx-1$

and $5>e$

$\displaystyle\lim_{x \to \infty}\frac{5^x}{1-lnx}=-\lim_{x \to \infty}\frac{5^x}{lnx-1}=-\lim_{x \to \infty}\frac{\left[\frac{5}{e}\right]^xe^x}{lnx-1}$

goes to negative infinity.
• February 21st 2011, 07:08 AM
Moklin
My teacher solved it from me but i think he made a calculus mistake :/
He said the answer was +oo

Anyway...can anyone solve this?

limx--0+ f(x)

f(x)= xln(x)

Without using L´Hopital rule.
• February 22nd 2011, 05:58 AM
Quote:

Originally Posted by Moklin
My teacher solved it from me but i think he made a calculus mistake :/
He said the answer was +oo

Anyway...can anyone solve this?

limx--0+ f(x)

f(x)= xln(x)

Without using L´Hopital rule.

Again, you could utilise the fact that

$e^x>log_ex$

Hence

$xe^x>xlog_ex$

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

the numerator approaches 1, while the denominator goes to infinity.

That limit goes to zero, in which case the other also goes to zero or a negative value.
Maybe you could examine that ?

Hint: You could compare the limits of $xlog_ex+1$ and $xe^x+1$