# Math Help - Getting the limit

1. ## 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).

2. $y=e^x,\;\;\;x=log_ey$

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

The functions differ by ever-increasing magnitude

3. Using a "series approach":

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

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

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

Another approach:

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

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

$\lim_{x\rightarrow\infty} e^x = \infty$ we get the desired result.

4. 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

5. 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.

6. 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

7. Also,

$e^x>2x^2lnx$

$\displaystyle\lim_{x \to \infty}\left(e^x-lnx\right)\;>\;\lim_{x \to \infty}\left(2x^2-1\right)lnx$

$lnx>1$ for $x>e$