A calculus problem by Guilherme Dela Corte

Calculus Level pending

lim ⁡ x → 2 ( 2 x 2 − 5 x + 2 ) cot ⁡ ( π x ) = a π b \displaystyle{\lim_{x \to 2 } } \left (2x^2-5x+2 \right ) \cot(\pi x) = a \pi ^b

On the limit above, a a and b b are integers. Evaluate − b π a -b \pi ^a to the nearest integer.


The answer is 31.

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Guilherme Dela Corte by Guilherme Dela Corte , 24, Brazil

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1 solution

From the start we have the indetermination lim ⁡ x → 2 ( 2 x 2 − 5 x + 2 ) cot ⁡ ( π x ) = 0 × ∞ \lim_{x \to 2} \left ( 2x^2-5x+2 \right ) \cot(\pi x) = 0 \times \infty . Rearranging the limit to lim ⁡ x → 2 2 x 2 − 5 x + 2 tan ⁡ ( π x ) = 0 0 \lim_{x \to 2} \frac{ 2x^2-5x+2 }{\tan(\pi x)} = \frac{0}{0} , we can apply L'Hôpital's Rule ( ★ ) ( \bigstar ) .

Doing so, we have:

lim ⁡ x → 2 2 x 2 − 5 x + 2 tan ⁡ ( π x ) = ★ lim ⁡ x → 2 4 x − 5 π sec ⁡ 2 ( π x ) = 3 π = 3 π − 1 . \lim_{x \to 2} \; \frac{ 2x^2-5x+2 }{\tan(\pi x)} \; \overset{\bigstar}{=} \; \lim_{x \to 2} \; \frac{4x-5}{\pi \sec^2(\pi x)} = \dfrac{3}{\pi} = 3\pi^{-1}.

Our desired value is − ( − 1 ) π 3 = 31.0062766803 ≈ 31. -(-1)\pi^3 = 31.0062766803 \approx \boxed{31.}

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