Question #91010

Question

1383 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-03T23:15:26

#lim_(x->0)(tan(x)-x)/(xsin(x))=0#

Direct substitution of #x=0# shows that the initial limit results in a #0/0# indeterminate form, and thus we can apply L'hopital's rule.

#lim_(x->0)(tan(x)-x)/(xsin(x)) = lim_(x->0)(d/dx(tan(x)-x))/(d/dxxsin(x))#

#=lim_(x->0)(sec^2(x)-1)/(xcos(x)+sin(x))#

As this again results in a #0/0# indeterminate form, we may apply L'hopital's rule once again:

#=lim_(x->0)(d/dx(sec^2(x)-1))/(d/dx(xcos(x)+sin(x)))#

#=lim_(x->0)(2sec^2(x)tan(x))/(2cos(x)-xsin(x))#

#=(2sec^2(0)tan(0))/(2cos(0)-0sin(0))#

#=(2*1*0)/(2*1-0*0)#

#=0#