Manipulative limit

Calculus Level 3

lim ⁡ x → 0 ( e x 2 − cos ⁡ ( 5 x ) x sin ⁡ ( x ) ) \large \displaystyle\lim_{x\to0} \left( \dfrac{e^{x^2}-\sqrt{\cos{(5x)}}}{x\sin{(x)}}\right)

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


The answer is 33.

Miloje Đukanović by Miloje Đukanović , 24, Serbia

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2 solutions

L = lim ⁡ x → 0 ( e x 2 − 1 + 1 − cos ⁡ 5 x x sin ⁡ x ) = lim ⁡ x → 0 ( e x 2 − 1 x sin ⁡ x ) + lim ⁡ x → 0 ( 1 − cos ⁡ 5 x x sin ⁡ x ) L= \displaystyle\lim_{x\to0} \left( \frac{e^{x^2}-1+1-\sqrt{\cos{5x}}}{x\sin{x}}\right)=\displaystyle\lim_{x\to0} \left( \frac{e^{x^2}-1}{x\sin{x}}\right)+\displaystyle\lim_{x\to0} \left( \frac{1-\sqrt{\cos{5x}}}{x\sin{x}}\right)

L = lim ⁡ x → 0 ( e x 2 − 1 x 2 sin ⁡ x x ) + lim ⁡ x → 0 ( 1 − cos ⁡ 5 x x sin ⁡ x 1 + cos ⁡ 5 x 1 + cos ⁡ 5 x ) L=\displaystyle\lim_{x\to0} \left( \frac{e^{x^2}-1}{x^2\frac{\sin{x}}{x}}\right)+\displaystyle\lim_{x\to0} \left( \frac{1-\sqrt{\cos{5x}}}{x\sin{x}}\color{#D61F06}{\frac{1+\sqrt{\cos{5x}}}{1+\sqrt{\cos{5x}}}}\right)

L = lim ⁡ x → 0 ( e x 2 − 1 x 2 x sin ⁡ x ) + lim ⁡ x → 0 ( 25 1 − cos ⁡ 5 x ( 5 x ) 2 x sin ⁡ x 1 1 + cos ⁡ 5 x ) = 1 ⋅ 1 + 25 ⋅ 1 2 ⋅ 1 ⋅ 1 2 = 29 4 L=\displaystyle\lim_{x\to0} \left( \frac{e^{x^2}-1}{x^2}\frac{x}{\sin{x}}\right)+\displaystyle\lim_{x\to0} \left( 25\frac{1-\cos{5x}}{(5x)^2}\frac{x}{\sin{x}}\frac{1}{1+\sqrt{\cos{5x}}}\right)=1\cdot1+25\cdot\frac{1}{2}\cdot1\cdot\frac{1}{2}=\frac{29}{4}

6 Helpful 0 Interesting 0 Brilliant 0 Confused
Prasad Mahadik
Nov 9, 2015

It's much more easier if binomial expansion to (cos5x)^.5 is applied for 1-25x^2/2

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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