Question

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

Answer 1

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`

Answer 2

`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`.