Simple Limit

Calculus Level 3

Find

lim x 0 e x 2 x e x x sin ( x ) \large \lim_{x \to 0} \dfrac{e^x-2x-e^{-x}}{x-\sin(x)}

e e 1 1 3 3 2 2

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1 solution

Chew-Seong Cheong
Mar 15, 2019

L = lim x 0 e x 2 x e x x sin x By Maclaurin series = lim x 0 ( 1 + x + x 2 2 ! + x 3 3 ! + ) 2 x ( 1 x + x 2 2 ! x 3 3 ! + ) x ( x x 3 3 ! + x 5 5 ! + ) = lim x 0 2 ( x 3 3 ! + x 5 5 ! + x 7 7 ! + ) x 3 3 ! + x 5 5 ! + x 7 7 ! + = 2 \begin{aligned} L & = \lim_{x \to 0} \frac {e^x-2x-e^{-x}}{x-\sin x} & \small \color{#3D99F6} \text{By Maclaurin series} \\ & = \lim_{x \to 0} \frac {\left(1+x + \frac {x^2}{2!} + \frac {x^3}{3!} + \cdots \right)-2x-\left(1-x + \frac {x^2}{2!} - \frac {x^3}{3!} + \cdots \right)}{x-\left(x - \frac {x^3}{3!} + \frac {x^5}{5!} + \cdots \right)} \\ & = \lim_{x \to 0} \frac {2\left(\frac {x^3}{3!} + \frac {x^5}{5!} + \frac {x^7}{7!} + \cdots \right)}{\frac {x^3}{3!} + \frac {x^5}{5!} + \frac {x^7}{7!} + \cdots} \\ & = \boxed 2 \end{aligned}


Reference: Maclaurin series

Beautiful way! I just more brute forcing: used l'hopital 3 times to get the answer.

Peter van der Linden - 2 years, 2 months ago

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