limit, x->0, [x - ln(1+x)] / [e^{3x} - x - 1] : Can it be done?

[unlucky]

determine the limit: lim x→0

[math]lim \; x→0 \; (x − ln(1 + x)) / (e^{3x} – x – 1)[/math]
can it be done? and how?

 

[BigBeachBanana]

its e^3x I don't know how to format it correctly

[math]\lim_{x \to 0} \dfrac{x−\ln(1+x)}{e^{3x}−x−1}[/math]Is this correct? If so, I see it's a candidate for L'Hopital.

 

[Steven G]

Can it be done? You decide! Let x be -.0009, evaluate what you are taking the limit of, then plug in for x say x=.0003 and evaluate what you are taking the limit of. Are these two results close to one another? If yes, then (most probably) the limit exists and if not, then the limit fails to exist.

If you some reason you don't think that your two results are close enough to decide that the limit exists, then pick two closer values to 0 (one less than 0 and one greater than 0) for x.

If you conclude that the limit exists, then and only then you can try to figure out the limit with confidence that it exists.

 

[MathTutorPro]

I would recommend doing L'Hopital's rule twice for this problem. After the first time that it's applied, you still get an indeterminate form. With using the rule multiple times, the only term that will remain in the denominator is the exponential term and exp(0) = 1 so it's no longer indeterminate.

 

[blamocur]

I would recommend doing L'Hopital's rule twice for this problem.

Using it once worked for me.

 

[lookagain]

I would recommend doing L'Hopital's rule twice for this problem. After the first time that it's applied, you still get an indeterminate form.

You cannot use it after the first time, because you will not get an indeterminate form after a first use.