y = 5 + x 2 3 − x d d x ( 5 + x 2 ) = 2 x 2 5 + x 2 = x 5 + x 2 d d x ( 3 − x ) = − 1 lim x → ± ∞ 5 + x 2 3 − x = lim x → ± ∞ − x 5 + x 2 (1), iif limit exists = lim x → ± ∞ ∓ x 2 5 + x 2 Remember negative values; alternatively, consider the positive and negative cases separately = lim x → ± ∞ ∓ 5 + x 2 − 5 5 + x 2 = lim x → ± ∞ ∓ 1 − 5 5 + x 2 → ∓ 1 Also satisfies (1) y = 5 + x 2 3 − x ⋮ x = 3 y 2 ± 14 y 2 − 5 y 2 − 1 = 3 ( y 2 − 1 ) + 3 ± 14 y 2 − 5 y 2 − 1 = 3 + ± 14 y 2 − 5 y 2 − 1 = 3 ± 14 y 2 − 5 ( y 2 − 1 ) 2 = 3 ± 14 z − 5 ( z − 1 ) 2 z = y 2 ∃ constants A , B . 14 z − 5 ( z − 1 ) 2 = A ( z − 1 ) 2 + B z − 1 Partial fractions lim z → ± ∞ 14 z − 5 ( z − 1 ) 2 = lim z → ± ∞ A ( z − 1 ) 2 + lim z → ± ∞ B z − 1 Limit of sums is sum of limits = 0 ∴ lim y → ± ∞ x = lim y → ± ∞ 3 ± 0 Note that z ≥ 0 = 3 {\displaystyle {\begin{aligned}y&=\displaystyle {\frac {\sqrt {5+x^{2}}}{3-x}}\\\displaystyle {\frac {d}{dx}}\left({\sqrt {5+x^{2}}}\right)&=\displaystyle {\frac {2x}{2{\sqrt {5+x^{2}}}}}\\&=\displaystyle {\frac {x}{\sqrt {5+x^{2}}}}\\\displaystyle {\frac {d}{dx}}(3-x)&=-1\\\displaystyle \lim _{x\to \pm \infty }\displaystyle {\frac {\sqrt {5+x^{2}}}{3-x}}&=\displaystyle \lim _{x\to \pm \infty }-\displaystyle {\frac {x}{\sqrt {5+x^{2}}}}&&{\mbox{(1), iif limit exists}}\\&=\displaystyle \lim _{x\to \pm \infty }\mp {\sqrt {\displaystyle {\frac {x^{2}}{5+x^{2}}}}}&&{\mbox{Remember negative values; alternatively, consider the positive and negative cases separately}}\\&=\displaystyle \lim _{x\to \pm \infty }\mp {\sqrt {\displaystyle {\frac {5+x^{2}-5}{5+x^{2}}}}}\\&=\displaystyle \lim _{x\to \pm \infty }\mp {\sqrt {1-\displaystyle {\frac {5}{5+x^{2}}}}}\\&\to \mp 1&&{\mbox{Also satisfies (1)}}\\y&=\displaystyle {\frac {\sqrt {5+x^{2}}}{3-x}}\\&\vdots \\x&=\displaystyle {\frac {3y^{2}\pm {\sqrt {14y^{2}-5}}}{y^{2}-1}}\\&=\displaystyle {\frac {3(y^{2}-1)+3\pm {\sqrt {14y^{2}-5}}}{y^{2}-1}}\\&=3+\displaystyle {\frac {\pm {\sqrt {14y^{2}-5}}}{y^{2}-1}}\\&=3\pm {\sqrt {\displaystyle {\frac {14y^{2}-5}{\left(y^{2}-1\right)^{2}}}}}\\&=3\pm {\sqrt {\displaystyle {\frac {14z-5}{\left(z-1\right)^{2}}}}}&&z=y^{2}\\\exists {\mbox{constants }}A,B\,.\,\\\displaystyle {\frac {14z-5}{\left(z-1\right)^{2}}}&=\displaystyle {\frac {A}{\left(z-1\right)^{2}}}+\displaystyle {\frac {B}{z-1}}&&{\mbox{Partial fractions}}\\\displaystyle \lim _{z\to \pm \infty }\displaystyle {\frac {14z-5}{\left(z-1\right)^{2}}}&=\displaystyle \lim _{z\to \pm \infty }\displaystyle {\frac {A}{\left(z-1\right)^{2}}}+\displaystyle \lim _{z\to \pm \infty }\displaystyle {\frac {B}{z-1}}&&{\mbox{Limit of sums is sum of limits}}\\&=0\\\therefore \displaystyle \lim _{y\to \pm \infty }x&=\displaystyle \lim _{y\to \pm \infty }3\pm {\sqrt {0}}&&{\mbox{Note that }}z\geq 0\\&=3\end{aligned}}}
