Limit Revision

Calculus Level 4

lim ⁡ x → 0 ( 1 sin ⁡ 2 x − 1 x 2 ) \displaystyle\lim_{x \rightarrow 0} \left( \dfrac{1}{\sin^2x} - \dfrac{1}{x^2}\right)

If the limit above is equal to a b \dfrac ab , where a a and b b are coprime positive integers, find the value of a + b a+b .

Submit your answer as 0 if you don't think the limit exist.


The answer is 4.

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Akhil Bansal by Akhil Bansal , 22, India

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2 solutions

The limit can be rewritten as lim ⁡ x → 0 ( x 2 − sin ⁡ 2 ( x ) x 2 sin ⁡ 2 ( x ) ) . \lim_{x \rightarrow 0} \left(\dfrac{x^{2} - \sin^{2}(x)}{x^{2}\sin^{2}(x)}\right).

Now in series form sin ⁡ ( x ) = x − x 3 6 + O ( x 5 ) , \sin(x) = x - \dfrac{x^{3}}{6} + O(x^{5}), giving us that sin ⁡ 2 ( x ) = x 2 − x 4 3 + O ( x 6 ) . \sin^{2}(x) = x^{2} - \dfrac{x^{4}}{3} + O(x^{6}).

Thus x 2 − sin ⁡ 2 ( x ) = x 4 3 + O ( x 6 ) x^{2} - \sin^{2}(x) = \dfrac{x^{4}}{3} + O(x^{6}) and x 2 sin ⁡ 2 ( x ) = x 4 + O ( x 6 ) , x^{2}\sin^{2}(x) = x^{4} + O(x^{6}), and so the limit becomes

lim ⁡ x → 0 x 4 3 + O ( x 6 ) x 4 + O ( x 6 ) = lim ⁡ x → 0 1 3 + O ( x 2 ) 1 + O ( x 2 ) = 1 3 . \lim_{x \rightarrow 0} \dfrac{\dfrac{x^{4}}{3} + O(x^{6})}{x^{4} + O(x^{6})} = \lim_{x \rightarrow 0} \dfrac{\dfrac{1}{3} + O(x^{2})}{1 + O(x^{2})} = \dfrac{1}{3}.

Thus a + b = 1 + 3 = 4 . a + b = 1 + 3 = \boxed{4}.

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From th expansion of sin ⁡ x \sin x , how did you write expansion of sin ⁡ 2 x \sin^2x ?

Akhil Bansal - 5 years, 7 months ago

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Expanding sin ⁡ 2 ( x ) , \sin^{2}(x), we have that

sin ⁡ 2 ( x ) = sin ⁡ ( x ) sin ⁡ ( x ) = ( x − x 3 6 + O ( x 5 ) ) ( x − x 3 6 + O ( x 5 ) ) = \sin^{2}(x) = \sin(x)\sin(x) = \left(x - \dfrac{x^{3}}{6} + O(x^{5})\right)\left(x - \dfrac{x^{3}}{6} + O(x^{5})\right) =

x 2 − 2 x ∗ x 3 6 + O ( x 6 ) = x 2 − x 4 6 + O ( x 6 ) . x^{2} - 2x*\dfrac{x^{3}}{6} + O(x^{6}) = x^{2} - \dfrac{x^{4}}{6} + O(x^{6}).

One could also note that

1 sin ⁡ 2 ( x ) − 1 x 2 = ( x + sin ⁡ ( x ) x sin ⁡ ( x ) ) ( x − sin ⁡ ( x ) x sin ⁡ ( x ) ) = ( 2 x + O ( x 3 ) x 2 + O ( x 4 ) ) ( x 3 6 + O ( x 5 ) x 2 + O ( x 4 ) ) , \dfrac{1}{\sin^{2}(x)} - \dfrac{1}{x^{2}} = \left(\dfrac{x + \sin(x)}{x\sin(x)}\right)\left(\dfrac{x - \sin(x)}{x\sin(x)}\right)= \left(\dfrac{2x + O(x^{3})}{x^{2} + O(x^{4})}\right)\left(\dfrac{\dfrac{x^{3}}{6} + O(x^{5})}{x^{2} + O(x^{4})}\right),

which goes to 2 6 = 1 3 \dfrac{2}{6} = \dfrac{1}{3} in the limit as x → 0. x \rightarrow 0.

Brian Charlesworth - 5 years, 7 months ago

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Thanks sir.. I got it now, i have one query...
As we know when x → 0 , sin ⁡ ( x ) → x ⇒ sin ⁡ 2 ( x ) → x 2 x \rightarrow 0 , \sin(x) \rightarrow x \Rightarrow \sin^2(x) \rightarrow x^2 .

Then, lim ⁡ x → 0 ( 1 sin ⁡ 2 x − 1 x 2 ) = lim ⁡ x → 0 ( 1 x 2 − 1 x 2 ) = 0 \displaystyle \lim_{x \rightarrow 0} \left( \dfrac{1}{\sin^2 x} - \dfrac{1}{x^2}\right) = \displaystyle \lim_{x \rightarrow 0} \left( \dfrac{1}{x^2} - \dfrac{1}{x^2}\right) = 0 .

What's flaw in that?

Akhil Bansal - 5 years, 7 months ago

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@Akhil Bansal – The concern here is that, while sin ⁡ 2 ( x ) \sin^{2}(x) does indeed go to x 2 x^{2} as x → 0 , x \rightarrow 0, the limit we are dealing with is of the indeterminate form ∞ − ∞ , \infty - \infty, so we can't take the direct approach you mention. This is why I chose to expand sin ⁡ ( x ) \sin(x) in its series form, so that we could better evaluate the behavior of this limit as x → 0. x \rightarrow 0.

Brian Charlesworth - 5 years, 7 months ago

Apply the L'Hopital's rule for lim ⁡ x → 0 ( 1 sin ⁡ x + 1 x 2 ) \lim_{x \to 0} \left( \frac{1}{\sin^x}+\frac{1}{x^2}\right)

ADIOS!!! \LARGE \text{ADIOS!!!}

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