Limit: $\displaystyle \lim _{x\to\infty}{\frac{e^x-3}{e^x+1}}$

Question

Is it correct to write $\displaystyle \lim _{x\to\infty}{\frac{e^x-3}{e^x+1}}=\lim _{x\to\infty}{\frac{e^x}{e^x}}=1$? Can I simply discard $-3$ and $+1$?

Answer 1

No, but you can multiply numerator and denominator by $e^{-x}$ :

$$\frac{e^x-3}{e^x+1}=\frac{1-3\, e^{-x}}{1+ e^{-x}}\to \frac{1+0}{1+0}=1$$

Answer 2

You can either compute it using the limit rule or you can use Squeeze theorem to prove it.

$$1-\frac{4}{e^x}\le 1-\frac{4}{e^x+1}=\frac{e^x-3}{e^x+1}\le 1$$

Answer 3

This is true

$$\lim _{x\to\infty}{\frac{e^x-3}{e^x+1}}=\lim _{x\to\infty}{\frac{e^x}{e^x}}=1$$

but this is not a correct way to proceed.

Usually, and more in general, we proceed factoring out the dominating term as follows

$$\frac{e^x-3}{e^x+1}=\frac{e^x}{e^x}\cdot \frac{1-\frac{3}{e^x}}{1+\frac{1}{e^x}}\to 1\cdot \frac{1-0}{1+0}=1$$

similarly to the example already given.

In this case, as a simple alternative, we can also proceed as follows

$$\frac{e^x-3}{e^x+1}=\frac{e^x+1-4}{e^x+1}=\frac{e^x+1}{e^x+1}-\frac{4}{e^x+1}=1-\frac{4}{e^x+1}\to 1-0=1$$

Answer 4

I think first you would want to prove that you can discard the -3 in the numerator, which you can do simply splitting the fraction in two, e^x/(1+e^x)-3/(1+e^x). Clearly the minus part goes to 0 as x goes to infinite and you're left with e^x/(1+e^x). That goes to 1, but you can prove it by multiplying top and bottom by e^-x to get
1/(e^(-x)+1)