Limits

**roshanhero** · October 4th 2008, 04:38 AM

lim.....asinx-sin2x/tan^3x is finite,find the value of 'a' and find the limit.
x->0

**galactus** · October 4th 2008, 06:39 AM

We can rewrite as:

$\lim_{x\to 0}\frac{(a-2cos(x))cos^{3}(x)}{sin^{2}(x)}$

Look at $a-2cos(x)$

What if a=2, what is the limit?.

**skeeter** · October 4th 2008, 06:46 AM

I'm assuming that this is your limit expression ...

$\lim_{x \to 0} \frac{a\sin{x} - \sin(2x)}{\tan^3{x}}$

using L'Hopital ...

$\lim_{x \to 0} \frac{a\cos{x} - 2\cos(2x)}{3\tan^2{x}\sec^2{x}}$

at this point, substituting 0 in the denominator yields 0 and substituting 0 in the numerator yields (a - 2). for the limit to exist, a must equal 2.

starting from scratch with a = 2 ...

$\lim_{x \to 0} \frac{2\sin{x} - \sin(2x)}{\tan^3{x}}$

$\lim_{x \to 0} \frac{2\sin{x} - 2\sin{x}\cos{x}}{\tan{x}(\sec^2{x} - 1)}$

$\lim_{x \to 0} \frac{2\sin{x}(1 - \cos{x})}{\tan{x}(\sec^2{x} - 1)}$

$\lim_{x \to 0} \frac{2\sin{x}(1 - \cos{x})}{\tan{x}(\sec{x}+1)(\sec{x}-1)}$

$\lim_{x \to 0} \frac{2\sin{x}(1 - \cos{x})}{\frac{\sin{x}}{\cos{x}}(\sec{x}+1)\left( \frac{1-\cos{x}}{\cos{x}}\right)}$

$\lim_{x \to 0} \frac{2}{\sec^2{x}(\sec{x}+1)} = 1$

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