Can we solve it?

Calculus Level 4

$\large \lim_{x\to0} \dfrac{ae^x-b\cos{x}+ce^{-x} }{x\sin{x}} = 2$

Find $a^3 \times b^4 \times c^5$

Clarification : $e$ denotes Euler's number , $e \approx 2.71828$ .

None of these choices 8 12 32 16 0

1 solution

展豪張
May 21, 2016

First we try to evaluate the LHS directly: it is $\dfrac{a-b+c}{0}$
For the limit to exists, let $a-b+c=0$ and use L'Hopital's rule.
The limit became $\dfrac{ae^x+b\sin x-ce^{-x}}{\sin x+x\cos x}$
Perform the same trick again: let $a=c$ and use L'Hopital's rule.
The limit became $\dfrac{ae^x+b\cos x+ce^{-x}}{2\cos x-x\sin x}$
The denominator is finally not $0$ : it is $2$
So we have $a+b+c=4$
Solving the above and we have $a=1,b=2,c=1$
Btw, it is a very crazy question...I love it (+1) =D

nice solution.. +1

Sabhrant Sachan - 5 years ago

Woah! And I thought that the tedious Maclaurin series is the best approach. +1

Pi Han Goh - 5 years ago

Can we solve it?

1 solution

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