# Calculate the limit: $\lim_{x\to \infty} (1-\frac{2x+1}{2x-1})^x$ [closed]

how can I solve this limit without using l'hopital.

$$\lim_{x\to \infty} \left(1-\frac{2x+1}{2x-1}\right)^x$$

• The problem is ill-posed. The base is always negative when $x > 0$. Commented Dec 25, 2017 at 21:55
• @Zhanxiong Well, no: the problem isn't wrong. This would be an excellent Calculus I question for the students to pay attention first of all to what the expression whose limit is asked really is...In this case, the expression is ill-defined...but the good student must come up with that! Commented Dec 25, 2017 at 21:58
• The expression simplifies to $(\frac{-2}{2x-1})^x$. The limit looks like $0^{\infty}$ Commented Dec 25, 2017 at 22:00
• Point of terminology: You ask, "how can I solve this integral?" There is no integral here. I think you mean, "how can I evaluate (not solve) this limit?" Commented Dec 25, 2017 at 22:01
• Ah! Yes, I just noticed. My apologies. Commented Dec 25, 2017 at 22:03

$$\left(1-\frac{2x+1}{2x-1}\right)^x=\left(-\frac2{2x-1}\right)^x=\ldots$$

the expression isn't well defined for infinite values of $\;x\;$ when $\;x \to\infty\;$ and thus the limit cannot be taken.

• Why can not I take positive values?. Thanks for the help. Commented Dec 25, 2017 at 22:07
• I think it's because it's an oscillating limit, right? Commented Dec 25, 2017 at 22:09
• @trick15f Because $\;-2<0\;$ whereas $\;2x+1>0\;$ since $\;x\to\infty\;$ and we can assume $\;x>0\;$ all the time... Commented Dec 25, 2017 at 22:12
• @trick15f Nothing oscillating here: the expression whose limit you wanted is not well defined when $\;x\to\infty\;$ as the basis of that power is negative . Commented Dec 25, 2017 at 22:16

$$\left(1-\frac{2x+1}{2x-1}\right)^x=\left(\frac{-2}{2x-1}\right)^x$$

thus the limit exist only for $x\to -\infty$, set $y=-x\to +\infty$

$$\left(\frac{-2}{2x-1}\right)^x=\left(\frac{2}{2y+1}\right)^{-y}=e^{-y\log{\frac{2}{2y+1}}}\to +\infty$$

• Didn't you read the other (my) answer and/or the comments below the question? Commented Dec 25, 2017 at 22:04
• @DonAntonio yes I've read your interpretation of the OP. I guess that from your point of view the limit for $x \to \infty$ should be interpreted in a rescritive way as a limit for $x \to +\infty$ and $x \to -\infty$?
– user
Commented Dec 25, 2017 at 22:11
• I'm not sure what you're asking: the limit when $\;x\to\infty\;$ means that the given function of $\;x\;$ must be evaluated when larger and larger (positive) values of $\;x\;$ are taken. What has $\;-\infty\;$ to do with this? Commented Dec 25, 2017 at 22:14
• @DonAntonio sometimes I've seen the notation $x\to \infty$ for $|x|\to +\infty$
– user
Commented Dec 25, 2017 at 22:23
• I've never seen such a thing, but even if it existed I presume we should be warned and correspondingly said by whoever uses such a bizarre convention. Commented Dec 25, 2017 at 22:26