Evaluate #lim_(x rarr oo) (2x)/3 * sin(3/x )* sec(5/x)# ?

2 Answers
Apr 28, 2018

the answer of #lim_(xrarroo)[(2x)/3*sin(3/x)*sec(5/x)]=2#

Explanation:

if your question was ypu must use laplace law

#lim_(xrarroo)[(2x)/3*sin(3/x)*sec(5/x)]# Direct compensation product #(oo*0)#

#lim_(xrarroo)[(sin(3/x)*sec(5/x))/(3/(2x)]]# Direct compensation product #(0/0)#

#f(x)=sin(3/x)*sec(5/x)#

#g(x)=(3/(2x))#

#lim_(xrarra)[(f(x)')/(g(x)')]# if the Direct compensation product equal #(0/0)#

#lim_(xrarroo)[[-(sec(5/x)*(5*sin(3/x)*tan(5/x)+3*cos(3/x)))/x^2]/[-3/(2*x^2)]]#

#lim_(xrarroo)[(2*(sec(5/x))*(5*sin(3/x)*tan(5/x)+3*cos(3/x)))/3]#

#=[[(2)(5*0*0+3))/3]=6/3=2#

Apr 28, 2018

# 2#.

Explanation:

We will use

#lim_(theta to 0)sintheta/theta=1, lim_(theta to 0)sectheta=1#.

Set #1/x=y," so that, as "x to oo, y to 0#.

#:." The Reqd. Lim."=lim_(y to 0)2/(3y)*sin3ysec5y#,

#=lim_((3y) to 0)2{(sin3y)/(3y)}{lim_((5y) to 0)sec5y}#,

#=2*1*1#,

#=2#.