Expert Community at Your Service
Solve your problem for the price of one coffee
Let f(x)=g(x)h(x),where g(x)=sinx−x.cosx,andh(x)=x3.The limit is x→0.Since g(0) and h(0) both =0,then use l'Hopital's rule, limf(x)[x−>0]=lim[x−>0]g″(x)/h″(x)g′(x)=cosx−cosx+x.sinx=x.sinxh′(x)=3x2Since g'(0) and h'(0) both = 0,then again use l'Hopital's rule, i.e. limf(x)[x−>0]=lim[x−>0]g″(x)/h″(x)g″(x)=sinx+x.cosxh″(x)=6xSince g''(0) and h''(0) both = 0,then again use l'Hopital's rule, i.e. limf(x)[x−>0]=lim[x−>0]g‴(x)/h‴(x)g‴(x)=cosx+cosx−x.sinx=2cosx−x.sinxh‴(x)=6Now, g″(0)=2, and h‴(0)=6.So, limf(x)[x−>0]=lim[x−>0]g‴(x)/h‴(x)=lim[x−>0]2/6=1/3
Ask your question. Get your answer. Easy as that
Get answers within minutes and finish your homework faster
Or
Dont have an account? Register
Create a free account to see answers
Already have an account? Sign in