Looking for familiar limits

Calculus Level 2

If L = lim ⁡ x → 0 tan ⁡ x − sin ⁡ x x 3 L = \displaystyle \lim_{x \rightarrow 0} \dfrac{\tan x - \sin x}{x^{3}} , find 10 L 10 L .

Credit: Piskunov's Differential and Integral Calculus


The answer is 5.

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Hobart Pao by Hobart Pao , 23, USA

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3 solutions

Rishabh Jain
Feb 6, 2016

U s e tan ⁡ x = sin ⁡ x cos ⁡ x and take sin x common . \small{\color{#000fff}{Use~\tan x=\dfrac{\sin x}{\cos x}\text{ and take sin x common .}}} = lim ⁡ x → 0 sin ⁡ x ( 1 − cos ⁡ x ) x 3 =\lim_{x\rightarrow0} \dfrac{\sin x(1-\cos x)}{x^3} = lim ⁡ x → 0 sin ⁡ x x × lim ⁡ x → 0 1 − cos ⁡ x x 2 =\lim_{x\rightarrow0} \dfrac{\sin x}{x}\times \lim_{x\rightarrow0} \dfrac{1-\cos x}{x^2} = 1 × 1 2 = 1 2 =\large 1\times \dfrac{1}{2}=\dfrac{1}{2} = 10 × 1 2 = 5 =\Large 10\times\dfrac{1}{2}=\boxed{\color{#302B94}{5}} lim ⁡ x → 0 sin ⁡ x x = 1 a n d lim ⁡ x → 0 1 − cos ⁡ x x 2 = 1 2 \color{#20A900}{\boxed{\color{#D61F06}{\lim_{x\rightarrow0} \dfrac{\sin x}{x}=1~and~\lim_{x\rightarrow0} \dfrac{1-\cos x}{x^2}=\dfrac{1}{2}}}}

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I guess that the first numerator should be tan ⁡ ( x ) ( 1 − cos ⁡ ( x ) ) \tan(x)(1 - \cos(x)) , which will still work since

lim ⁡ x → 0 tan ⁡ ( x ) x = 1 \displaystyle\lim_{x \to 0} \dfrac{\tan(x)}{x} = 1 as well. To prove the second "red" limit for others, note that

lim ⁡ x → 0 1 − cos ⁡ ( x ) x 2 = lim ⁡ x → 0 ( 1 − cos ⁡ ( x ) x 2 ∗ 1 + cos ⁡ ( x ) 1 + cos ⁡ ( x ) ) = lim ⁡ x → 0 sin ⁡ 2 ( x ) x 2 ( 1 + cos ⁡ ( x ) ) = \displaystyle\lim_{x \to 0} \dfrac{1 - \cos(x)}{x^{2}} = \lim_{x \to 0} \left(\dfrac{1 - \cos(x)}{x^{2}} * \dfrac{1 + \cos(x)}{1 + \cos(x)}\right) = \lim_{x \to 0} \dfrac{\sin^{2}(x)}{x^{2}(1 + \cos(x))} =

( lim ⁡ x → 0 sin ⁡ ( x ) x ) 2 ∗ lim ⁡ x → 0 1 1 + cos ⁡ ( x ) = 1 2 ∗ 1 1 + 1 = 1 2 . \displaystyle\left(\lim_{x \to 0} \dfrac{\sin(x)}{x}\right)^{2} * \lim_{x \to 0} \dfrac{1}{1 + \cos(x)} = 1^{2} * \dfrac{1}{1 + 1} = \dfrac{1}{2}.

Brian Charlesworth - 5 years, 4 months ago

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Also like this:

lim ⁡ x → 0 1 − cos ⁡ ( x ) x 2 = lim ⁡ x → 0 4 ( 1 − cos ⁡ ( x ) ) 4 x 2 = lim ⁡ x → 0 sin ⁡ 2 ( x / 2 ) 2 × x 2 4 = 1 2 lim ⁡ x → 0 sin ⁡ ( x / 2 ) x 2 × lim ⁡ x → 0 sin ⁡ ( x / 2 ) x 2 = 1 2 × 1 × 1 = 1 2 \begin{aligned} \lim_{x\rightarrow 0} \frac{1-\cos(x)}{x^2} &= \lim_{x\rightarrow 0} \frac{4(1-\cos(x))}{4x^2} \\ &=\lim_{x\rightarrow 0} \frac{\sin^2(x/2)}{2\times \frac{x^2}{4}} \\ &=\frac{1}{2}\lim_{x\rightarrow 0} \frac{\sin(x/2)}{\frac{x}{2}} \times \lim_{x\rightarrow 0} \frac{\sin(x/2)}{\frac{x}{2}}\\ &=\frac{1}{2} \times 1 \times 1 = \boxed{\dfrac{1}{2}} \end{aligned}

PS: Trigo is cool :P

Nihar Mahajan - 5 years, 4 months ago

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One more method is to use Maclaurin series , i.e.,

tan ⁡ ( x ) = x + x 3 3 + O ( x 5 ) , sin ⁡ ( x ) = x − x 3 6 + O ( x 5 ) , \tan(x) = x + \dfrac{x^{3}}{3} + O(x^{5}), \sin(x) = x - \dfrac{x^{3}}{6} + O(x^{5}),

which means that tan ⁡ ( x ) − sin ⁡ ( x ) = x 3 2 + O ( x 5 ) . \tan(x) - \sin(x) = \dfrac{x^{3}}{2} + O(x^{5}).

Dividing through by x 3 x^{3} gives us lim ⁡ x → 0 ( 1 2 + O ( x 2 ) ) = 1 2 \displaystyle\lim_{x \to 0}\left(\dfrac{1}{2} + O(x^{2})\right) = \dfrac{1}{2} .

Brian Charlesworth - 5 years, 4 months ago

That is cool. I haven't seen it done that way before. :)

Brian Charlesworth - 5 years, 4 months ago

I took the LCM and placed cos0=1. = lim ⁡ x → 0 sin ⁡ x x 3 ( 1 cos ⁡ x − 1 ) = lim ⁡ x → 0 sin ⁡ x x 3 ( 1 − cos ⁡ x 1 ) . . . . . . since lim ⁡ x → 0 cos ⁡ x = 1 =\lim_{x\rightarrow0}\dfrac{\sin x}{x^3} ( \dfrac{1}{\cos x}-1)\\ =\lim_{x\rightarrow0} \dfrac{\sin x}{x^3} (\dfrac{1-\cos x}{1})......\text{since}\lim_{x\rightarrow0} \cos x=1 Anyways, nicely explained..

Rishabh Jain - 5 years, 4 months ago

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Ah, o.k.. That looks good, then. :)

Brian Charlesworth - 5 years, 4 months ago
Department 8
Feb 7, 2016

As x approaches the value pf the expression approaches 0.5

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Atul Shivam
Feb 6, 2016

Hint:- apply L'hospital rure till 0 0 \frac{0}{0} vanishes and then put limit which will gives L = 0.5 L=0.5 but we have to find 10 L 10L which is 5 \boxed{5}

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I personally found L'Hopital's for this question quite tedious...

Nihar Mahajan - 5 years, 4 months ago

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It's not bad when you consider that tan(0)=0 and sec(0)=1, since the only somewhat annoying derivatives that arise involve sec(x) and tan(x). With that in mind, you can ignore most of the details in calculating the third derivative of tan(x)

Tristan Goodman - 1 year, 7 months ago

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