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

Question

6066 views around the world

You can reuse this answer
Creative Commons License

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#