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

$2$

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

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

$(x^2-5x+2)/(x+4) = (x^2+4x-9x+2)/(x+4) = (x^2+4x-9x-36+38)/(x+4)$

$= (x(x+4)-9(x+4)+38)/(x+4) = x-9+38/(x+4)$

to see that

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

Now we have

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

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

As $xrarr0$ , we see that $(x^2+4x)rarr0$ so we get:

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

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

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