Close to cosine

Calculus Level 3

8 lim ⁡ x → 2 1 − cos ⁡ ( x 3 − 6 x 2 + 11 x − 6 ) ( x − 2 ) 2 = ? \large 8 \lim_{x\to2} \dfrac{1 - \cos(x^3-6x^2 + 11x-6)}{(x-2)^2} = \, ?


The answer is 4.

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Rudraksh Sisodia by Rudraksh Sisodia , 23, India

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1 solution

Sabhrant Sachan
Aug 24, 2016

x 3 − 6 x 2 + 11 x − 6 = ( x − 1 ) ( x − 2 ) ( x − 3 ) 1 − cos ⁡ ( x 3 − 6 x 2 + 11 x − 6 ) = 2 sin ⁡ 2 ( ( x − 1 ) ( x − 2 ) ( x − 3 ) 2 ) From the above equations Limit = 16 lim ⁡ x → 2 sin ⁡ 2 ( ( x − 1 ) ( x − 2 ) ( x − 3 ) 2 ) ( x − 1 ) 2 ( x − 2 ) 2 ( x − 3 ) 2 4 ⋅ ( x − 1 ) 2 ( x − 3 ) 2 4 Limit = 16 × 1 4 = 4 x^3-6x^2+11x-6 = (x-1)(x-2)(x-3) \\ 1-\cos{(x^3-6x^2+11x-6)} = 2\sin^2{\left(\dfrac{(x-1)(x-2)(x-3)}{2}\right)} \\ \text{From the above equations }\\ \text{Limit = } 16\displaystyle\lim_{x \to 2}\dfrac{\sin^2{\left(\dfrac{(x-1)(x-2)(x-3)}{2}\right)}}{\dfrac{(x-1)^2(x-2)^2(x-3)^2}{4}} \cdot \dfrac{(x-1)^2(x-3)^2}{4} \\ \text{Limit = } 16 \times \dfrac{1}{4} =\boxed{ 4 }

5 Helpful 0 Interesting 0 Brilliant 0 Confused

Wow that substitution with the concept of lim(x->0) sin(x)/x! So darn cool.

Noah Hunter - 4 years, 9 months ago

@Sambhrant Sachan yeah thats a good one !! +1 ,,

Rudraksh Sisodia - 4 years, 9 months ago

Cool substitution, but it doesn't really exploitatie why that limit equals 1/4. You stil het 0/0 if you would fill in x=2. So basically the question was rewritten. I used l'Hopitals rules twice to get 8 x 1/2 = 4

Peter van der Linden - 4 years, 9 months ago

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