`lim_(x->oo) (int_1^x ln(e^(4t-1)) dt )/ x` Evaluate the limit, using L’Hôpital’s Rule if necessary.

`lim_(x->oo) (int_1^x ln(e^(4t-1)) dt )/ x`

= `lim_(x->oo) (int_1^x (4t-1) dt )/ x`

= `lim_(x->oo) ( [4(t^2)/2-t]_1^x )/ x`

= `lim_(x->oo) ( [2x^2 -x]-[2(1^2)-1] )/ x`

= ` lim_(x->oo) ( [2x^2 -x]-[2-1] ) /x`

= `lim_(x->oo) ( [2x^2 -x-1 ]) /x`

= `lim_(x->oo) [2x -1-1/x ]`

upon plugging the value of `x = oo` , then we get

`lim_(x->oo) [2x -1-1/x ]`

= `[2(oo) -1-1/(oo) ]`

= `2(oo) -1 -0`

= ` oo`

Comments

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