What is the limit as n approaches infinity of sqrt(n^4+2n^3)-(n^2+n) ?

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limit as n approaches infinity of sqrt(n^4+2n^3)-(n^2+n)

n \to \infty lim (n^{4} + 2 n^{3} - (n^{2} + n))

- \frac{1}{2}

Decimal Notation

- 0.5

Solution steps

n \to \infty lim (n^{4} + 2 n^{3} - (n^{2} + n))

Multiply by the conjugate of

n^{4} + 2 n^{3} - (n^{2} + n) : \frac{- n ^{2}}{n ^{4} + 2 n ^{3} + ( n ^{2} + n )}

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

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

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

Divide by highest denominator power:

\frac{1}{1 + \frac{2}{n} + 1 + \frac{1}{n}}

= - n \to \infty lim \frac{1}{1 + \frac{2}{n} + 1 + \frac{1}{n}}

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 _{n \to \infty} ( 1 )}{lim _{n \to \infty} ( 1 + \frac{2}{n} + 1 + \frac{1}{n} )}

n \to \infty lim (1) = 1

n \to \infty lim (1 + \frac{2}{n} + 1 + \frac{1}{n}) = 2

= - \frac{1}{2}

\frac{8 x}{e ^{x}}

\int_{0}^{5} 5 x - 1 d x

\int_{0}^{1} (9 x^{e} + 2 e^{x}) d x

\frac{d y}{d x} = 2 x y^{2} + 3 x^{2} y^{2}

f (t) = \frac{cos ( t )}{t ^{7}}

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