What is the limit as h approaches infinity of (3h+2xh^2+x^2+h^3)/(4-3xh-2x^3h^3) ?

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limit as h approaches infinity of (3h+2xh^2+x^2+h^3)/(4-3xh-2x^3h^3)

h \to \infty lim (\frac{3 h + 2 x h ^{2} + x ^{2} + h ^{3}}{4 - 3 x h - 2 x ^{3} h ^{3}})

- \frac{1}{2 x ^{3}}

Solution steps

h \to \infty lim (\frac{3 h + 2 x h ^{2} + x ^{2} + h ^{3}}{4 - 3 x h - 2 x ^{3} h ^{3}})

Divide by highest denominator power:

\frac{\frac{3}{h ^{2}} + \frac{2 x}{h} + \frac{x ^{2}}{h ^{3}} + 1}{\frac{4}{h ^{3}} - \frac{3 x}{h ^{2}} - 2 x ^{3}}

= h \to \infty lim (\frac{\frac{3}{h ^{2}} + \frac{2 x}{h} + \frac{x ^{2}}{h ^{3}} + 1}{\frac{4}{h ^{3}} - \frac{3 x}{h ^{2}} - 2 x ^{3}})

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 _{h \to \infty} ( \frac{3}{h ^{2}} + \frac{2 x}{h} + \frac{x ^{2}}{h ^{3}} + 1 )}{lim _{h \to \infty} ( \frac{4}{h ^{3}} - \frac{3 x}{h ^{2}} - 2 x ^{3} )}

h \to \infty lim (\frac{3}{h ^{2}} + \frac{2 x}{h} + \frac{x ^{2}}{h ^{3}} + 1) = 1

h \to \infty lim (\frac{4}{h ^{3}} - \frac{3 x}{h ^{2}} - 2 x^{3}) = - 2 x^{3}

= \frac{1}{- 2 x ^{3}}

Simplify

= - \frac{1}{2 x ^{3}}

\frac{d}{d x} (e^{6 x s i n (2 x)})

x \to 3 lim (sin (\frac{1}{x - 3}))

\frac{d}{d x} ((x^{2} + 5)^{\frac{1}{2}})

cos^{4} (t) - sin^{4} (t)

y^{^{'}} + 3 y = 18 x

What is the limit as h approaches infinity of (3h+2xh^2+x^2+h^3)/(4-3xh-2x^3h^3) ?
The limit as h approaches infinity of (3h+2xh^2+x^2+h^3)/(4-3xh-2x^3h^3) is -1/(2x^3)