I am trying to solve this limit:
limx→∞2xx200
The first result is an indetermination of the kind ∞∞ but here applying L'Hopital would be too long, I do not see any substitution by means of equivalent infinitesimals possible and simplifying the limit neither.
How can I solve it? The solution must be ∞. Thank you!
Answer
L'Hopital rule applied 200 times leads to
limx→∞(log2)2002x200!=∞
Or you can take the logarithm of the limit
loglimx→∞2xx200=limx→∞(xlog2−200logx)=limx→∞x(log2−200logxx)=∞
As the log of the limit is +∞ the limit is +∞
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