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Calculus Level 4

lim ⁡ α → 0 e cos ⁡ ( α n ) − e α m = − e 2 \large \lim_{\alpha\rightarrow 0}\frac{e^{\cos(\alpha^{n})}-e}{\alpha^{m}} = -\frac{e}{2}

Let m m and n n be two constant positive integers greater than 1 such that the equation above holds true.

Find the ratio m n \frac{m}{n} .


The answer is 2.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

Kushal Bose
Jan 10, 2017

e lim ⁡ x → 0 e cos ⁡ ( x n ) − 1 − 1 x m = e lim ⁡ x → 0 e cos ⁡ ( x n ) − 1 − 1 cos ⁡ ( x n ) − 1 × cos ⁡ ( x n ) − 1 x m = e × 1 × lim ⁡ x → 0 cos ⁡ ( x n ) − 1 x m = − e × lim ⁡ x → 0 2 sin ⁡ 2 ( x n 2 ) ( x n 2 ) 2 × ( x n / 2 ) 2 x m = − e 2 × lim ⁡ x → 0 x 2 n − m \displaystyle e \, \lim_{x \to 0} \dfrac{e^{\cos (x^n)-1}-1}{x^m} \\ =e \, \displaystyle \lim_{x \to 0} \dfrac{e^{\cos (x^n)-1}-1}{\cos (x^n)-1} \times \dfrac{\cos (x^n)-1}{x^m} \\ = e \displaystyle \times 1 \times \lim_{x \to 0} \dfrac{\cos (x^n)-1}{x^m} \\ =-e \times \displaystyle \lim_{x \to 0} \dfrac{2 \sin^2 (\frac{x^n}{2})}{(\frac{x^n}{2})^2} \times \dfrac{(x^{n}/2)^2}{x^m} \\ =\frac{-e}{2} \times \displaystyle \lim_{x \to 0}{x^{2n-m}}

So, 2 n − m = 0 ⟹ 2 n = m 2n-m=0 \implies 2n=m

So the ratio m n \dfrac{m}{n} is 2 2

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is this from jee advanced ' 15?

Kunal Gupta - 4 years, 4 months ago

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