Question

Find limit x tends to 0 3sinx-sin3X/X3

Answer 1

Answer:

Step-by-step explanation:

= lim 3sinx - sin3x/x^3

x _0

= lim x tends to 0

3sinx - 3sinx + 4sin^3x/x^3

= lim 4sin^3x/x^3

x_0

= 4

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Answer 2

The value of the limit is 4.

Step-by-step explanation:

The given limit is

$\lim _{x\to \:0}\left(\frac{3\sin \left(x\right)-\sin \left(3x\right)}{x^3}\right)$

Apply L' Hospital's rule

$\lim _{x\to \:0}\left(\frac{3\cos \left(x\right)-\cos \left(3x\right)\cdot \:3}{3x^2}\right)\\\\=\lim _{x\to \:0}\left(\frac{\cos \left(x\right)-\cos \left(3x\right)}{x^2}\right)$

Again apply L' Hospital's rule

$\lim _{x\to \:0}\left(\frac{-\sin \left(x\right)+3\sin \left(3x\right)}{2x}\right)$

Again apply L' Hospital's rule

$\lim _{x\to \:0}\left(\frac{-\cos \left(x\right)+9\cos \left(3x\right)}{2}\right)$

Plugging the value x = 0

$\frac{-\cos \left(0\right)+9\cos \left(3\cdot \:0\right)}{2}\\\\=\frac{-1+9}{2}\\\\=\frac{8}{2}\\\\=4$

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