What is the limit as y approaches infinity of (2y^2-3y+5)/(y^2-5y+2) ?

The limit as y approaches infinity of (2y^2-3y+5)/(y^2-5y+2) is 2

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limit as y approaches infinity of (2y^2-3y+5)/(y^2-5y+2)

y \to \infty lim (\frac{2 y ^{2} - 3 y + 5}{y ^{2} - 5 y + 2})

2

Solution steps

y \to \infty lim (\frac{2 y ^{2} - 3 y + 5}{y ^{2} - 5 y + 2})

Divide by highest denominator power:

\frac{2 - \frac{3}{y} + \frac{5}{y ^{2}}}{1 - \frac{5}{y} + \frac{2}{y ^{2}}}

= y \to \infty lim (\frac{2 - \frac{3}{y} + \frac{5}{y ^{2}}}{1 - \frac{5}{y} + \frac{2}{y ^{2}}})

x \to a lim [\frac{f ( x )}{g ( x )}] = \frac{lim _{x \to a} f ( x )}{lim _{x \to a} g ( x )}, x \to a lim g (x) \neq = 0

With the exception of indeterminate form

= \frac{lim _{y \to \infty} ( 2 - \frac{3}{y} + \frac{5}{y ^{2}} )}{lim _{y \to \infty} ( 1 - \frac{5}{y} + \frac{2}{y ^{2}} )}

y \to \infty lim (2 - \frac{3}{y} + \frac{5}{y ^{2}}) = 2

y \to \infty lim (1 - \frac{5}{y} + \frac{2}{y ^{2}}) = 1

= \frac{2}{1}

Simplify

= 2

x \to 2 lim (x^{3} + 3 x^{2})

f (x) = 8 x^{2} - x^{3}

\int x^{3} x^{2} + 17 d x

\frac{d v}{d t} = k \cdot (275 - v), v (0) = 0

cos^{2} (x) + sin^{2} (x)

What is the limit as y approaches infinity of (2y^2-3y+5)/(y^2-5y+2) ?
The limit as y approaches infinity of (2y^2-3y+5)/(y^2-5y+2) is 2