Parametric Relations

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as $x\rightarrow \infty \quad we\quad have\quad t\rightarrow \frac { \pi }{ 2 }$

Clearly $y\rightarrow 3$

Hence $\lim _{ x\rightarrow \infty }{ y } =3$

$y=3{ sin }^{ 2 }t\\ x=tan(t)\\ t={ tan }^{ -1 }x\\ therefore,\quad y=3{ (sin{ tan }^{ -1 }x) }^{ 2 }\\ y=3({ sin{ sin }^{ -1 }\frac { x }{ \sqrt { 1+{ x }^{ 2 } } } })^{ 2 }\\ y=\frac { 3{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \\ \underset { x\rightarrow \infty }{ lim } y=3$

$y = 3 sin^2 t = 3 (1-cos^2 t) = 3 (1-\frac{1}{1-sec^2 t}) = 3 (1-\frac{1}{1+tan^2 t})$

let: $u = tan t$

so: $lim_{x\to\infty} y = lim_{u\to\infty}3(1-\frac{1}{1+u^2}) = \boxed{3}$

As x approaches infinity, $\tan(t)$ approaches infinity. We know that as t approaches 90º, $\tan(90º)=infinity$

Therefore when $x$ approaches infinity, $t=90º$ . Substituting 90º in y, we get

$y=3sin^2(90º)$ = $3*1^2$ =3