Find the value of limit sin(x^2+4x)/(x^3-5x^2+2x) as x approaches to 0?

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Find the value of limit sin(x^2+4x)/(x^3-5x^2+2x) as x approaches to 0?

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Answer 1 · 2017-12-06T20:37:19

$2$ $2$

Explanation:

Note that: $\frac{x^{3} - 5 x^{2} + 2 x}{x^{2} + 4 x} = \frac{x (x^{2} - 5 x + 2)}{x (x + 4)}$ $(x^3-5x^2+2x)/(x^2+4x) = (x(x^2-5x+2))/(x(x+4))$ .

Do the division: $\frac{x^{2} - 5 x + 2}{x + 4}$ $(x^2-5x+2)/(x+4)$ .

$\frac{x^{2} - 5 x + 2}{x + 4} = \frac{x^{2} + 4 x - 9 x + 2}{x + 4} = \frac{x^{2} + 4 x - 9 x - 36 + 38}{x + 4}$ $(x^2-5x+2)/(x+4) = (x^2+4x-9x+2)/(x+4) = (x^2+4x-9x-36+38)/(x+4)$

$= \frac{x (x + 4) - 9 (x + 4) + 38}{x + 4} = x - 9 + \frac{38}{x + 4}$ $= (x(x+4)-9(x+4)+38)/(x+4) = x-9+38/(x+4)$

to see that

$(x^{3} - 5 x^{2} + 2 x) = x (x + 4) (x - 9 + \frac{38}{x + 4})$ $(x^3-5x^2+2x) = x(x+4)(x-9+38/(x+4))$ .

Now we have

$\frac{sin (x^{2} + 4 x)}{x^{3} - 5 x^{2} + 2 x} = \frac{sin (x^{2} + 4 x)}{(x^{2} + 4 x) (x - 9 + \frac{38}{x + 4})}$ $sin(x^2+4x)/(x^3-5x^2+2x) = sin(x^2+4x)/((x^2+4x)(x-9+38/(x+4))$

$= \frac{sin (x^{2} + 4 x)}{(x^{2} + 4 x)} \cdot \frac{1}{x - 9 + \frac{38}{x + 4}}$ $= sin(x^2+4x)/((x^2+4x)) * 1/(x-9+38/(x+4))$

As $x \to 0$ $xrarr0$ , we see that $(x^{2} + 4 x) \to 0$ $(x^2+4x)rarr0$ so we get:

$lim_(xrarr0)sin(x^2+4x)/(x^3-5x^2+2x) = lim_(xrarr0)sin(x^2+4x)/((x^2+4x)) * lim_(xrarr0)1/(x-9+38/(x+4))$

$= 1 * 1/(0-9+38/(0+4))$

$= 1/(-9+17/2) = 1/(1/2) = 2$

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Find the value of limit sin(x^2+4x)/(x^3-5x^2+2x) as x approaches to 0?

1 Answer

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