How do you find the limit of #(xcos(ax))/(tan(bx))# as x approaches 0?

Question

2666 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-10T12:31:42

Limit of #(xcos(ax))/tan(bx)#, as #x# approaches #0# is #1/b#

#(xcos(ax))/tan(bx)=(xcos(ax))/(sin(bx)/cos(bx))#

or #(xcos(ax)cos(bx))/sin(bx)#

Hence #L_(x->0)(xcos(ax))/tan(bx)=L_(x->0)(xcos(ax)cos(bx))/sin(bx)#

or #L_(x->0)1/bxx(bx)/sin(bx)xxL_(x->0)cos(ax)xxL_(x->0)cos(bx)#

but as #x->0#, #cospx->1# and #L_(z->0)z/sinz=1#,

the above is equal to

#1/bxxL_(x->0)(bx)/sin(bx)xxL_(x->0)cos(ax)xxL_(x->0)cos(bx)#

or #1/bxx1xx1xx1=1/b#