Maclaurin series?

Calculus Level 3

L = lim ⁡ x → 0 e x a x − ( a x + e x ) + ln ⁡ ( 1 + x ) e x − 1 − e x sin ⁡ x + sin ⁡ x ( 1 + 2 e x sin ⁡ x ) + e x cos ⁡ ( 2 x ) e 2 x − 1 L = \lim_{x\to0} \dfrac{e^x a^x-(a^x+e^x)+\ln{(1+x)^{e^{x}-1}} -e^x\sin{x}+\sin x(1+2e^x\sin{x})+e^x\cos{(2x)} }{e^{2x}-1}

Let a a be a constant positive real number , find the value of L L .

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

None of these choices 2 Limit does not exist − 1 2 -\frac{1}{2} 0 0 1 2 \frac{1}{2}

Sabhrant Sachan by Sabhrant Sachan , 21, India

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1 solution

Chew-Seong Cheong
May 31, 2016

L = lim ⁡ x → 0 e x a x − ( a x + e x ) + ln ⁡ ( 1 + x ) e x − 1 − e x sin ⁡ x + sin ⁡ x ( 1 + 2 e x sin ⁡ x ) + e x cos ⁡ ( 2 x ) e 2 x − 1 = lim ⁡ x → 0 e x a x − a x − e x + ln ⁡ ( 1 + x ) e x − 1 − e x sin ⁡ x + sin ⁡ x + 2 e x sin ⁡ 2 x + e x cos ⁡ ( 2 x ) e 2 x − 1 = lim ⁡ x → 0 ( e x − 1 ) a x + ( e x − 1 ) ln ⁡ ( 1 + x ) − ( e x − 1 ) sin ⁡ x − e x ( 1 − 2 sin ⁡ 2 x ) + e x cos ⁡ ( 2 x ) ( e x − 1 ) ( e x + 1 ) = lim ⁡ x → 0 ( e x − 1 ) a x + ( e x − 1 ) ln ⁡ ( 1 + x ) − ( e x − 1 ) sin ⁡ x − e x cos ⁡ ( 2 x ) + e x cos ⁡ ( 2 x ) ( e x − 1 ) ( e x + 1 ) = lim ⁡ x → 0 a x + ln ⁡ ( 1 + x ) − sin ⁡ x e x + 1 = 1 + 0 − 0 1 + 1 = 1 2 \begin{aligned} L & = \lim_{x\to0} \frac{e^x a^x-(a^x+e^x)+\ln{(1+x)^{e^{x}-1}} -e^x\sin{x}+\sin x(1+2e^x\sin{x})+e^x\cos{(2x)} }{e^{2x}-1} \\ & = \lim_{x\to 0} \frac{e^x a^x-a^x\color{#D61F06}{-e^x}+\ln (1+x)^{e^{x}-1} -e^x\sin{x}+\sin x\color{#D61F06}{+2e^x\sin^2{x}}+e^x\cos{(2x)}}{e^{2x}-1} \\ & = \lim_{x\to 0} \frac{(e^x -1) a^x+(e^x -1) \ln (1+x) - (e^x-1) \sin{x}\color{#D61F06}{- e^x(1-2\sin^2{x})}+e^x\cos{(2x)}}{(e^x-1)(e^x+1)} \\ & = \lim_{x\to 0} \frac{(e^x -1) a^x+(e^x -1) \ln (1+x) - (e^x-1) \sin{x}\color{#D61F06}{- e^x \cos (2x)}+e^x\cos{(2x)}}{(e^x-1)(e^x+1)} \\ & = \lim_{x\to 0} \frac{a^x+\ln (1+x) - \sin{x}}{e^x+1} \\ & = \frac{1+0-0}{1+1} \\ & = \boxed{\dfrac{1}{2}} \end{aligned}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Let function in numerator be f ( x ) f(x) and in denominator be g ( x ) g(x) .

By using L'Hospital rule :

Now L i m i t r e q = L t x − > 0 f ( x ) / g ( x ) = L t x − > 0 f ′ ( x ) / g ′ ( x ) Limit_{req} = Lt_{x-> 0 } f(x)/g(x) = Lt_{x->0} f^{'}(x)/g^{'}(x) .

Since this argumant of limit is not of form 0/0 , we find

L r e q = f ′ ( 0 ) / g ′ ( 0 ) = 0.5 L_{req}= f^{'}(0)/g^{'}(0) = \boxed{0.5} .

Aakash Khandelwal - 5 years ago

In the third line it should be (1-2(sinx)^2) Probably you forgot to remove e^x :)

Yuri Lombardo - 4 years, 10 months ago

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Thanks. I have changed it.

Chew-Seong Cheong - 4 years, 10 months ago

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