Find the following limit
limx→01x3((2+cosx3)x−1)
without using L'Hopital's rule.
I tried to solve this using fundamental limits such as limx→0(1+x)1x=e,limx→0xx=1 and equivalent infinitesimals at x→0 such as x∼sinx,ax∼1+xlna,(1+x)a∼1+ax. This is what I did so far:
limx→01x3((2+cosx3)x−1)=limx→0(3−(1−cosx)3)x−1x3=limx→0(3−12x23)x−1x3=limx→0(1−16x2)x−1x3
I used fact that x∼sinx at x→0. After that, I tried to simplify (1−16x2)x. Problem is because this is not indeterminate, so I cannot use infinitesimals here. Can this be solved algebraically using fundamental limits, or I need different approach?
Answer
You are certainly on the right track (although you've used an approximation that needs justification). Now write (1−x2/6)x=exp(xln(1−x2/6)). More approximations: ln(1−h)≈−h and eh≈1+h for h small. What happens if each ≈ is an =? That will tell you what the limit is. Now you need to make sure these approximations really work.
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