Evaluate the limit

Calculus Level 2

lim ⁡ x → 0 sin ⁡ 2 x − sin ⁡ 4 x s i n 4 x − sin ⁡ 8 x \lim _{ x\xrightarrow { } 0 }{ } \cfrac { \sin\quad 2x\quad -\quad \sin\quad 4x\quad }{ sin\quad 4x\quad -\quad \sin\quad 8x }


The answer is 0.5.

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2 solutions

Here is an approach that does not require L'Hopital's rule.

Using the identity sin ⁡ ( 2 A ) = 2 sin ⁡ ( A ) cos ⁡ ( A ) \sin(2A) = 2\sin(A)\cos(A) , the limit becomes

lim ⁡ x → 0 sin ⁡ ( 2 x ) ( 1 − 2 cos ⁡ ( 2 x ) ) sin ⁡ ( 4 x ) ( 1 − 2 cos ⁡ ( 4 x ) ) = lim ⁡ x → 0 sin ⁡ ( 2 x ) ( 1 − 2 cos ⁡ ( 2 x ) ) 2 sin ⁡ ( 2 x ) cos ⁡ ( 2 x ) ( 1 − 2 cos ⁡ ( 4 x ) ) = l i m x → 0 1 − 2 cos ⁡ ( 2 x ) 2 cos ⁡ ( 2 x ) ( 1 − 2 cos ⁡ ( 4 x ) ) \lim_{x \rightarrow 0} \dfrac{\sin(2x)(1 - 2\cos(2x))}{\sin(4x)(1 - 2\cos(4x))} = \lim_{x \rightarrow 0} \dfrac{\sin(2x)(1 - 2\cos(2x))}{2\sin(2x)\cos(2x)(1 - 2\cos(4x))} = lim_{x \rightarrow 0} \dfrac{1 - 2\cos(2x)}{2\cos(2x)(1 - 2\cos(4x))} .

Now we can just plug in x = 0 x = 0 to get an answer of 1 − 2 2 ∗ ( 1 − 2 ) = 1 2 = 0.5 \frac{1 - 2}{2*(1 - 2)} = \frac{1}{2} = \boxed{0.5} .

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I guess this is always the best approach, i. e., removing the indetermination by elementary algebra instead of using L'Hôpital. :)

Mikael Marcondes - 6 years, 2 months ago
Pravin Dharmaraj
Apr 21, 2015

Divide both numerator and denominator by x and use the formula.

So simple.

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