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1-6*x-x^2/3

1 - 6*x - x^2/3

1 - 6*x - x^(2/3)

-x^2 - 6*x + 1/3

1 - (6*x - x^2)/3

1 - 6*(x - x^2)/3

1 - 6*x - x^2/3

1 - 6*x - x*2/3

Limit of the function 1-6*x-x^2/3

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     /           2\
     |          x |
 lim |1 - 6*x - --|
x->oo\          3 /

\lim_{x \to \infty}\left(- \frac{x^{2}}{3} - 6 x + 1\right)

Limit(1 - 6*x - x^2/3, x, oo, dir='-')

Detail solution

Let's take the limit

\lim_{x \to \infty}\left(- \frac{x^{2}}{3} - 6 x + 1\right)

Let's divide numerator and denominator by x^2:

\lim_{x \to \infty}\left(- \frac{x^{2}}{3} - 6 x + 1\right)

\lim_{x \to \infty}\left(\frac{- \frac{1}{3} - \frac{6}{x} + \frac{1}{x^{2}}}{\frac{1}{x^{2}}}\right)

Do Replacement

u = \frac{1}{x}

then

\lim_{x \to \infty}\left(\frac{- \frac{1}{3} - \frac{6}{x} + \frac{1}{x^{2}}}{\frac{1}{x^{2}}}\right) = \lim_{u \to 0^+}\left(\frac{u^{2} - 6 u - \frac{1}{3}}{u^{2}}\right)

\frac{- \frac{1}{3} + 0^{2} - 0}{0} = -\infty

The final answer:

\lim_{x \to \infty}\left(- \frac{x^{2}}{3} - 6 x + 1\right) = -\infty

Lopital's rule

There is no sense to apply Lopital's rule to this function since there is no indeterminateness of 0/0 or oo/oo type

The graph

Plot the graph

Rapid solution [src]

-oo

-\infty

Expand and simplify

Other limits x→0, -oo, +oo, 1

\lim_{x \to \infty}\left(- \frac{x^{2}}{3} - 6 x + 1\right) = -\infty

\lim_{x \to 0^-}\left(- \frac{x^{2}}{3} - 6 x + 1\right) = 1

\lim_{x \to 0^+}\left(- \frac{x^{2}}{3} - 6 x + 1\right) = 1

\lim_{x \to 1^-}\left(- \frac{x^{2}}{3} - 6 x + 1\right) = - \frac{16}{3}

\lim_{x \to 1^+}\left(- \frac{x^{2}}{3} - 6 x + 1\right) = - \frac{16}{3}

\lim_{x \to -\infty}\left(- \frac{x^{2}}{3} - 6 x + 1\right) = -\infty

The graph