Limits 8

Calculus Level 3

Find lim ⁡ x → 0 1 − 2 sin ⁡ 2 x − cos ⁡ 3 x 5 x 2 \displaystyle\lim_{x \rightarrow 0} \displaystyle\frac{1-2\sin^{2}x-\cos^{3} x}{5x^{2}}


The answer is -0.1.

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Samuel Ayinde by samuel ayinde , 24, Nigeria

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2 solutions

Ahmed Essam
Apr 8, 2015

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Daniel Ferreira
Aug 1, 2015

lim ⁡ x → 0 1 − 2 ⋅ sin ⁡ 2 x − cos ⁡ 3 x 5 x 2 = lim ⁡ x → 0 1 − 2 ⋅ ( 1 − cos ⁡ 2 x ) − cos ⁡ 3 x 5 x 2 = lim ⁡ x → 0 2 ⋅ cos ⁡ 2 x − cos ⁡ 3 x − 1 5 x 2 = 0 0 ( indeterminate ) \\ \lim_{x \to 0} \frac{1 - 2 \cdot \sin^2 x - \cos^3 x}{5x^2} = \\\\\\ \lim_{x \to 0} \frac{1 - 2 \cdot (1 - \cos^2 x) - \cos^3 x}{5x^2} = \\\\\\ \lim_{x \to 0} \frac{2 \cdot \cos^2 x - \cos^3 x - 1}{5x^2} = \frac{0}{0} (\text{indeterminate})

Aplicando L'Hôspital,

lim ⁡ x → 0 4 ⋅ cos ⁡ x ⋅ ( − sin ⁡ x ) − 3 cos ⁡ 2 x ⋅ ( − sin ⁡ x ) 10 x = lim ⁡ x → 0 sin ⁡ x ⋅ ( − 4 ⋅ cos ⁡ x + 3 ⋅ cos ⁡ 2 x ) 10 x = lim ⁡ x → 0 sin ⁡ x x ⋅ − 4 ⋅ cos ⁡ x + 3 ⋅ cos ⁡ 2 x 10 \\ \lim_{x \to 0} \frac{4 \cdot \cos x \cdot (- \sin x) - 3\cos^2 x \cdot (- \sin x)}{10x} = \\\\\\ \lim_{x \to 0} \frac{\sin x \cdot (- 4 \cdot \cos x + 3 \cdot \cos^2 x)}{10x} = \\\\\\ \lim_{x \to 0} \frac{\sin x}{x} \cdot \frac{- 4 \cdot \cos x + 3 \cdot \cos^2 x}{10}

Do limite fundamental trigonométrico,

1 ⋅ − 4 ⋅ 1 + 3 ⋅ 1 10 = − 1 10 \\ 1 \cdot \frac{- 4 \cdot 1 + 3 \cdot 1}{10} = \\\\\\ \boxed{\boxed{- \frac{1}{10}}}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

okay i did it differently and got a wrong answer but i don't know why. So what i did is a broke 1-cos^3x and did difference of cubes. Then i used the identity that (1-cosx)/x=0 so that whole part became zero. That left me with -2sin^2x/5x^2 and i used the identity sinx/x=1 to get -.4

Ashish Sacheti - 5 years, 9 months ago

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