# Getting the limit

• Oct 15th 2010, 07:52 AM
wonderboy1953
Getting the limit
Reviewed my textbook on limits. It went over various types including the difference between limits that go to infinity. A problem occurred to me that the book didn't cover so I'm posting this as a puzzle.

Let's define this function: g(x) = exp(x) - ln(x) and let x approach infinity. What value does g(x) take on when x approaches infinity? (I'm wondering whether using the series approach will solve this problem).
• Oct 15th 2010, 12:02 PM
$\displaystyle y=e^x,\;\;\;x=log_ey$

Hence $\displaystyle y=log_ex$ is the mirror-image of $\displaystyle y=e^x$ through the line $\displaystyle y=x$

The functions differ by ever-increasing magnitude
• Oct 15th 2010, 02:50 PM
Unbeatable0
Using a "series approach":

$\displaystyle e^x \ge 1+x+\frac{x^2}{2}$ and $\displaystyle \ln x \le x+1$ (for $\displaystyle x\ge 1$) we get

$\displaystyle e^x-\ln x \ge (1+x+\frac{x^2}{2}) - (x+1) = \frac{x^2}{2}$

from which the limit follows, since $\displaystyle \lim_{x\rightarrow\infty} \frac{x^2}{2} = \infty$

Another approach:

$\displaystyle \lim_{x\rightarrow\infty} e^x-\ln x = \lim_{x\rightarrow\infty} (1-\frac{\ln x}{e^x}) e^x$

and since $\displaystyle \lim_{x\rightarrow\infty} 1-\frac{\ln x}{e^x} = 1$ and

$\displaystyle \lim_{x\rightarrow\infty} e^x = \infty$ we get the desired result.
• Oct 15th 2010, 10:38 PM
CaptainBlack
Quote:

Originally Posted by wonderboy1953
Reviewed my textbook on limits. It went over various types including the difference between limits that go to infinity. A problem occurred to me that the book didn't cover so I'm posting this as a puzzle.

Let's define this function: g(x) = exp(x) - ln(x) and let x approach infinity. What value does g(x) take on when x approaches infinity? (I'm wondering whether using the series approach will solve this problem).

Since exp(x) grows faster than any power of x, and log(x) grows more slowly than any non-zero power of x their difference grows faster than any power of x.

CB
• Oct 16th 2010, 08:30 AM
wonderboy1953
Quote:

Originally Posted by CaptainBlack
Since exp(x) grows faster than any power of x, and log(x) grows more slowly than any non-zero power of x their difference grows faster than any power of x.

CB

That's my line of reasoning CB. I was just looking for a more rigorous line of thought on what I suspected.
• Oct 16th 2010, 11:29 PM
CaptainBlack
Quote:

Originally Posted by wonderboy1953
That's my line of reasoning CB. I was just looking for a more rigorous line of thought on what I suspected.

It is actually rigorous, I was just expressing known big-O results in colloquial English.

CB
• Oct 17th 2010, 03:56 AM
$\displaystyle e^x>2x^2lnx$
$\displaystyle \displaystyle\lim_{x \to \infty}\left(e^x-lnx\right)\;>\;\lim_{x \to \infty}\left(2x^2-1\right)lnx$
$\displaystyle lnx>1$ for $\displaystyle x>e$