This is tricky.
Simplify #(sin(3t) + 4t) / (t sec(t))# into #(cos(t)*(sin(3t) + 4t)) / t#
Then apply some addition formulas and double-angle formulas.
Firstly let #sin(3t) = sin(2t + t)#
#= sin(2t)*cos(t) + cos(2t)*sin(t)#.
Then expand this with two compound angle formulas:
#sin(2t)*cos(t) + cos(2t)*sin(t)#
# = 2sin(t)*cos(t)*cos(t) + (2cos^2(t) - 1)*sin(t)#.
# = 2sin(t)*cos^2(t) +2cos^2(t)*sin(t) - sin(t)#
# = 4sin(t)*cos^2(t) - sin(t)#.
Then add the #4t#:
#4sin(t)*cos^2(t) - sin(t) + 4t#.
Then multiply by #cos(t)# so finally the original numerator is:
#4sin(t)*cos^3(t) - sin(t)*cos(t) + 4t*cos(t)#.
So, to find:
#lim_(t rarr 0) (4sin(t)*cos^3(t) - sin(t)*cos(t) + 4t*cos(t)) / t#
I shall steer clear of fancy formulas, and rely on an intuitive approach, if this suits.
Set the following values:
The approach of #t# to zero can be expressed as #t = 1/oo#.
Likewise, as #t# approaches zero, #sin(t) = 1/oo# and #cos(t) = 1#.
Hence, remove any #cos# terms from the expression.
So the numerator becomes:
#4sin(t)*cos^3(t) - sin(t)*cos(t) + 4t*cos(t)# #=4/oo - 1/oo + 4/oo#
#= 7/oo#
Finally, dividing by the #t# in the denominator:
#(7/oo)/ (1/oo) = 7#.
