`y = 3x - sin^2(x)` Find the limit, if possible

Textbook Question

Chapter 3, 3.9 - Problem 19 - Calculus of a Single Variable (10th Edition, Ron Larson).
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gsarora17 | (Level 2) Associate Educator

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`lim_(x->0)3x-sin^2x`

plug in the value of x

`= 3*0-sin^2(0)`

=0

`lim_(x->oo) 3x-sin^2x`

`lim_(x->oo) x(3-(sin^2x)/x)`

Apply the squeeze  theorem to evaluate the limit of sin^2x/x

`-1<=sinx<=1`

`0<=sin^2x<=1`

`lim_(x->oo)(0/x)<=lim_(x->oo)(sin^2x)/x<=lim_(x->oo)(1/x)`

`lim_(x->oo)(0/x)=0`

`lim_(x->oo)(1/x)=0`

So, by the squeeze theorem,

`lim_(x->oo)(sin^2x)/x=0`

Therefore,

`lim_(x->oo)3(x-(sin^2x)/x)=oo(3-0)=oo`

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