how can I solve this limit without using l'hopital.
$$\lim_{x\to \infty} \left(1-\frac{2x+1}{2x-1}\right)^x$$
8
-
2$\begingroup$ The problem is ill-posed. The base is always negative when $x > 0$. $\endgroup$– ZhanxiongCommented Dec 25, 2017 at 21:55
-
1$\begingroup$ @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! $\endgroup$– DonAntonioCommented Dec 25, 2017 at 21:58
-
$\begingroup$ The expression simplifies to $(\frac{-2}{2x-1})^x$. The limit looks like $0^{\infty}$ $\endgroup$– herb steinbergCommented Dec 25, 2017 at 22:00
-
2$\begingroup$ 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?" $\endgroup$– G Tony JacobsCommented Dec 25, 2017 at 22:01
-
$\begingroup$ Ah! Yes, I just noticed. My apologies. $\endgroup$– Naweed G. SeldonCommented Dec 25, 2017 at 22:03
|
Show 3 more comments
2 Answers
$\begingroup$
$\endgroup$
4
$$\left(1-\frac{2x+1}{2x-1}\right)^x=\left(-\frac2{2x-1}\right)^x=\ldots$$
the expression isn't well defined for infinite values of $\;x\;$ when $\;x \to\infty\;$ and thus the limit cannot be taken.
answered Dec 25, 2017 at 21:55
-
$\begingroup$ Why can not I take positive values?. Thanks for the help. $\endgroup$– trick15fCommented Dec 25, 2017 at 22:07
-
$\begingroup$ I think it's because it's an oscillating limit, right? $\endgroup$– trick15fCommented Dec 25, 2017 at 22:09
-
$\begingroup$ @trick15f Because $\;-2<0\;$ whereas $\;2x+1>0\;$ since $\;x\to\infty\;$ and we can assume $\;x>0\;$ all the time... $\endgroup$ Commented Dec 25, 2017 at 22:12
-
$\begingroup$ @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 . $\endgroup$ Commented Dec 25, 2017 at 22:16
$\begingroup$
$\endgroup$
8
$$ \left(1-\frac{2x+1}{2x-1}\right)^x=\left(\frac{-2}{2x-1}\right)^x$$
thus the limit exist only for $x\to -\infty$, set $y=-x\to +\infty$
$$\left(\frac{-2}{2x-1}\right)^x=\left(\frac{2}{2y+1}\right)^{-y}=e^{-y\log{\frac{2}{2y+1}}}\to +\infty$$
-
$\begingroup$ Didn't you read the other (my) answer and/or the comments below the question? $\endgroup$ Commented Dec 25, 2017 at 22:04
-
$\begingroup$ @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 $? $\endgroup$– userCommented Dec 25, 2017 at 22:11
-
$\begingroup$ 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? $\endgroup$ Commented Dec 25, 2017 at 22:14
-
$\begingroup$ @DonAntonio sometimes I've seen the notation $x\to \infty$ for $|x|\to +\infty$ $\endgroup$– userCommented Dec 25, 2017 at 22:23
-
1$\begingroup$ 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. $\endgroup$ Commented Dec 25, 2017 at 22:26