Please help evaluate the following expression. It seems that it is 0. But I have no clue how that happens.

(log is based on e. don't know about the constraints about a and b. Let's assume they are not 0s).

$$\lim_\infty { } b\cdot log(e^{ax}+1)-a\cdot (e^{bx}+1)$$

## A question about limits

### Re: A question about limits

I assume that the second term has a log in front of it as well?

The quickest solution is probably to observe that $$\lim_{x \rightarrow \infty} log (e^{ax} + 1 ) = ax$$, which shows that the limit you care about is zero.

Alternatively, you could start by rewriting the expression as the log of a fraction, interchange the limit and the log and then show that the limit of the fraction is 1:

$$\lim_{x \rightarrow \infty }log \frac{ (e^{ax}+1)^b}{(e^{bx}+1)^a }= log \lim_{x \rightarrow \infty } \frac{ (e^{ax}+1)^b}{(e^{bx}+1)^a }$$

The quickest solution is probably to observe that $$\lim_{x \rightarrow \infty} log (e^{ax} + 1 ) = ax$$, which shows that the limit you care about is zero.

Alternatively, you could start by rewriting the expression as the log of a fraction, interchange the limit and the log and then show that the limit of the fraction is 1:

$$\lim_{x \rightarrow \infty }log \frac{ (e^{ax}+1)^b}{(e^{bx}+1)^a }= log \lim_{x \rightarrow \infty } \frac{ (e^{ax}+1)^b}{(e^{bx}+1)^a }$$

### Re: A question about limits

Thanks for helping me out again, owlpride. I should have put a big parentheses for these two terms.owlpride wrote:I assume that the second term has a log in front of it as well?

The quickest solution is probably to observe that $$\lim_{x \rightarrow \infty} log (e^{ax} + 1 ) = ax$$, which shows that the limit you care about is zero.

Alternatively, you could start by rewriting the expression as the log of a fraction, interchange the limit and the log and then show that the limit of the fraction is 1:

$$\lim_{x \rightarrow \infty }log \frac{ (e^{ax}+1)^b}{(e^{bx}+1)^a }= log \lim_{x \rightarrow \infty } \frac{ (e^{ax}+1)^b}{(e^{bx}+1)^a }$$

The key lies in this thing $$\lim_{x \rightarrow \infty} log (e^{ax} + 1 ) = ax$$. Is this a common rule or something that obvious? That's where I got stuck. Could you also show some intermediate steps?

### Re: A question about limits

Sorry, my bad. There is another condition, a>0 and b>0. in this case, it's obvious $$\lim_{x \rightarrow \infty} log (e^{ax} + 1 ) = ax$$.Hom wrote:Thanks for helping me out again, owlpride. I should have put a big parentheses for these two terms.owlpride wrote:I assume that the second term has a log in front of it as well?

The quickest solution is probably to observe that $$\lim_{x \rightarrow \infty} log (e^{ax} + 1 ) = ax$$, which shows that the limit you care about is zero.

Alternatively, you could start by rewriting the expression as the log of a fraction, interchange the limit and the log and then show that the limit of the fraction is 1:

$$\lim_{x \rightarrow \infty }log \frac{ (e^{ax}+1)^b}{(e^{bx}+1)^a }= log \lim_{x \rightarrow \infty } \frac{ (e^{ax}+1)^b}{(e^{bx}+1)^a }$$

The key lies in this thing $$\lim_{x \rightarrow \infty} log (e^{ax} + 1 ) = ax$$. Is this a common rule or something that obvious? That's where I got stuck. Could you also show some intermediate steps?

### Re: A question about limits

I think that you meant lim ax, not ax