How can I solve this limit without using d'hospital method?

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SUMMARY

The limit \(\lim_{x\to0}\frac{\tan x-x}{x^3}\) can be solved without using L'Hôpital's rule by employing the Taylor series expansion for \(\tan(x)\). The series is given by \(\tan(x) = x + \frac{1}{3}x^3 + \frac{2}{15}x^5 + \frac{17}{315}x^7 + \ldots\). By substituting this expansion into the limit expression, the \(x\) terms cancel, allowing for simplification. After dividing both the numerator and denominator by \(x^3\) and applying direct substitution, the limit evaluates to \(\frac{1}{3}\).

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Homework Statement


hello, what can I do to solve this without using d'hospital method?
\lim_{x\to0}\frac{\tan x-x}{x^3}
 
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player1_1_1 said:

Homework Statement


hello, what can I do to solve this without using d'hospital method?
\lim_{x\to0}\frac{\tan x-x}{x^3}

Recall the formula [which is just the M.S. for tan(x)]:
tan(x)=x+\frac{1}{3}x^3+\frac{2}{15}x^5+\frac{17}{315}x^7+...
Substitute this in the limit and the two "x" will be cancelled.
Then divide the denominator and the numerator by x^3.
and the limit will be done by the direct substitution.
The limit = 1/3
 
look up the taylor series for sin and cosine and write the one for tan. Now factor out X^3 from the top and take that limit as X goes to Zero
 

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