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I want to calculate the following limit:

$$\lim_{x\to 0} (1+2x)^{2/(3x+1)} $$

It is clear to see that by plugging the value $x=0$ the result is 1. I want to calculate the limit using L'Hospital's rule, by applying $a^x=e^{x\ln(a)}.$ I got :

$$e^{2\ln(1+2x)/(3x+1)}$$

Then I tried to calculate the power limit by using l'Hospital's rule, and eventually calculate e^(power limit answer). still could not find the correct answer that way. Can somebody help me? Thanks in advance.

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  • $\begingroup$ The limit is 1. No l’Hospital needed. $\endgroup$ – gnasher729 Dec 5 '17 at 19:28
  • $\begingroup$ Thanks, I mentioned that but my task is solving it by using lh'opitals rules $\endgroup$ – user451803 Dec 5 '17 at 19:30
  • $\begingroup$ @user451803 This cannot be solved with L'Hospital's rule. $\endgroup$ – Rebellos Dec 5 '17 at 19:52
  • $\begingroup$ @user451803 If you are ok, you can set as solved. One of the right one I mean. Thanks! $\endgroup$ – user Dec 6 '17 at 13:31
  • $\begingroup$ @user451803 Please, if you are ok, you can accept the answer and set it as solved. Thanks! $\endgroup$ – user Jan 22 '18 at 21:32
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let $y$ = $(1+2x)^{2/(3x+1)}$

$ln$ $y$ = $\frac{2}{(3x+1)}$ $ln$ $(1+2x)$

taking the limit then take the Exponential $e$ of the answer

is that what you do ?!

because its not $\frac{0}{0}$ or $\frac{\infty}{\infty}$ the answer will be vary So ,

$\lim_{x\to0}$ $\frac{4}{3(2x+1)}$ apply L'H again to get $\lim_{x\to0}$ $\frac{0}{6}$ = 0

$e^{0}$ = 1

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  • $\begingroup$ Yeah exactly, Thank you for assisting me. Thanks everybody who answered on here. $\endgroup$ – user451803 Dec 5 '17 at 20:25
  • $\begingroup$ This is not correct on all cases. LHospitals rule is meant for undetermined forms. $\endgroup$ – Rebellos Dec 5 '17 at 20:29
  • $\begingroup$ But i think it's just a practical question to use the exponential Technic for solving other questions @Rebellos $\endgroup$ – sdfghjkiujyhtgrfdsxdcvbnhjkloi Dec 5 '17 at 20:39
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L'Hospital's rule works for limits of the form :

$$\frac{0}{0}$$

$$\frac{\infty}{\infty}$$

etc, which are called undetermined forms.

You cannot apply it on other cases, the limit of the derivatives of the denominator and numerator will not be the same as the limit of the initial expression.

As you've mentioned, this is a straight-forward limit to calculate, so :

$$\lim_{x\to 0} (1+2x)^{2/(3x+1)} = 1$$

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You can’t apply L'Hopital's rule as the form is not indeterminate.

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  • $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review $\endgroup$ – Badam Baplan Dec 5 '17 at 19:57
  • $\begingroup$ What are you talking about? He is asking literally: " I want to calculate the limit using L'Hospital's rule"! $\endgroup$ – user Dec 5 '17 at 20:01
  • $\begingroup$ Oh I didn't realize it leaves this generic comment behind! Next time I will post my own thoughts. $\endgroup$ – Badam Baplan Dec 5 '17 at 20:11
  • $\begingroup$ In my own approach to reviewing, I often flag answers that I believe should have been left instead as comments. Admittedly that's a sketchy MO; it's totally non-objective and objectionable. But this is community moderating! (And my word is small) $\endgroup$ – Badam Baplan Dec 5 '17 at 20:19
  • $\begingroup$ My reasoning here (on a question about popular knowledge) is that other users are inevitably going to post answers with the same message but including substantial additional explanation/clarification (e.g. what are the indeterminate forms to which L'Hospital's rule applies? Why doesn't the rule apply to other limit forms? etc.) I understand the drive to garner reputation, but sometimes I think it becomes distracting, and a comment, if anything, suffices. $\endgroup$ – Badam Baplan Dec 5 '17 at 20:23

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