I cannot figure this limit out.
limx→0+(ex−1)(tanx)23√x2
I've used the e to the ln trick and multiplied by 1 (x2x2) and arrived at
limx→0+exp(x4/3ln(ex−1))
However I failed at getting further. I tried adding and subtracting lnx but that got me nowhere.
I cannot use l'Hospital or Taylor series (only the "known" limits for sin, cos, ex, ln such as limx→0sinxx=1 which are really only Taylor series).
Thanks for help!
Answer
Hint: limx→0ex−1x=limx→0tanxx=1
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