What is the limit as w approaches infinity of (2w^2-3w+4)/(5w^2+7w-1) ?

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limit as w approaches infinity of (2w^2-3w+4)/(5w^2+7w-1)

w \to \infty lim (\frac{2 w ^{2} - 3 w + 4}{5 w ^{2} + 7 w - 1})

\frac{2}{5}

Decimal Notation

0.4

Solution steps

w \to \infty lim (\frac{2 w ^{2} - 3 w + 4}{5 w ^{2} + 7 w - 1})

Divide by highest denominator power:

\frac{2 - \frac{3}{w} + \frac{4}{w ^{2}}}{5 + \frac{7}{w} - \frac{1}{w ^{2}}}

= w \to \infty lim (\frac{2 - \frac{3}{w} + \frac{4}{w ^{2}}}{5 + \frac{7}{w} - \frac{1}{w ^{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 _{w \to \infty} ( 2 - \frac{3}{w} + \frac{4}{w ^{2}} )}{lim _{w \to \infty} ( 5 + \frac{7}{w} - \frac{1}{w ^{2}} )}

w \to \infty lim (2 - \frac{3}{w} + \frac{4}{w ^{2}}) = 2

w \to \infty lim (5 + \frac{7}{w} - \frac{1}{w ^{2}}) = 5

= \frac{2}{5}

y^{^{''}} - k y = 0

x \to - 2 lim ((x - 6)^{\frac{2}{3}})

y^{^{'}} = 0.2 ln (\frac{2000}{y}) y

\frac{d}{d x} (\frac{6}{1 + cos ( x )})

x \to \infty lim (0)

What is the limit as w approaches infinity of (2w^2-3w+4)/(5w^2+7w-1) ?
The limit as w approaches infinity of (2w^2-3w+4)/(5w^2+7w-1) is 2/5