I would like to evaluate this limit without using Taylor expansion:
limx→0ex−x−1x2
Note: by Taylor Expansion i have got :12 .
Thank u for any help .!!!!
Answer
METHODOLOGY 1: Use L'Hospital's Rule Successively
Repeated use of L'Hospital's Rule reveals
limx→0ex−x−1x2=limx→0ex−12x=limx→012ex=12
METHODOLOGY 2: Integral representation of the numerator
Note that we can write the numerator as
ex−x−1=∫x0∫t0esdsdt=∫x0∫xsesdtds=∫x0(x−s)esds
Now, we can use the Mean-Value-Theorem for integrals to reveal
ex−x−1=es∗(x)∫x0(x−s)ds=12x2es∗(x)
for some value of s∗(x)∈(0,x).
Finally, exploiting the continuity of the exponential function yield the coveted limit
limx→0ex−x−1x2=limx→012x2es∗(x)x2=12
as expected!
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