Limits by Taylor series
Pi Han Goh,
Prince Loomba,
Hjalmar Orellana Soto,
and
contributed
This wiki is incomplete.
This wiki is a stub. Help improve it.
When we want to evaluate a limit of a function, it is sometimes useful to know the Taylor series of the function itself, or an approximation if it is seasonable.
Prove
\[ \displaystyle \lim_{x\to0} \dfrac{\tan x - x}{x^3} = \dfrac13.\]
Knowing the Taylor series of \(\tan x\) centered at \(x=0 \) is \(x+ \frac13 x^3 + O\big(x^5\big) \), we have
\[\dfrac{\tan x - x}{x^3} = \dfrac{\frac13 x^3 + O\big(x^5\big) }{x^3} ={\dfrac13 x^2 + O\big(x^2\big)}. \]
Hence,
\[ \displaystyle \lim_{x\to0} \dfrac{\tan x - x}{x^3} = \lim_{x\to0} \left( \dfrac13 x^2 + O\big(x^2\big)\right) = \dfrac13.\ _\square\]
\[\lim_{x\to 0} \left(\dfrac{1}{x^2} - \dfrac{e^x}{(e^x-1)^2}\right)\]
Find the reciprocal of the above limit.
See Also
Cite as:
Limits by Taylor series.
Brilliant.org.
Retrieved from
https://brilliant.org/wiki/limits-by-taylor-series/