Ask your own question, for FREE!
Mathematics 19 Online
OpenStudy (anonymous):

lim x->infinity ((x^2)/(x+1))-((x^2)/(x-1))

OpenStudy (anonymous):

-2

OpenStudy (anonymous):

thanks. but i know it is -2, but i don't know how to set it up manually

OpenStudy (anonymous):

you can rewrite as \[\lim_{x \rightarrow \infty} 2x^4/(x-x^2)(x^2+x)\]

OpenStudy (anonymous):

then factor out the constants to get \[2(\lim_{x \rightarrow \infty}x^4/(x-x^2)(x^2+x))\]

OpenStudy (anonymous):

then use L'Hospitals rule to get \[2(\lim_{x \rightarrow \infty} dx^4/dx/d((x-x^2)(x+x^2))/dx)\]

OpenStudy (anonymous):

which is: \[2(\lim_{x \rightarrow \infty} 4x^3/((2x+1)(x-x^2) +(1-2x)(x^2+x))\]

OpenStudy (anonymous):

then factor out the constants again: \[8((\lim_{x \rightarrow \infty} x^3/((2x+1)(x-x^2)+(1-2x)(x^2+x))\]

OpenStudy (anonymous):

then use l'hospitals two more times to get -2

OpenStudy (anonymous):

thanks :)

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!