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Limit of the function (-1+x)/(x+x^2)

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     /-1 + x\
 lim |------|
x->1+|     2|
     \x + x /

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

Limit((-1 + x)/(x + x^2), x, 1)

Detail solution

Let's take the limit

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

transform

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

\lim_{x \to 1^+}\left(\frac{x - 1}{x \left(x + 1\right)}\right)

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

\frac{-1 + 1}{1 + 1} =

= 0

The final answer:

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

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

Other limits x→0, -oo, +oo, 1

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

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

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

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

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

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

Rapid solution [src]

0

Expand and simplify

One‐sided limits [src]

     /-1 + x\
 lim |------|
x->1+|     2|
     \x + x /

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

0

= 7.27864599832406e-29

     /-1 + x\
 lim |------|
x->1-|     2|
     \x + x /

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

0

= 6.05462511935933e-36

= 6.05462511935933e-36

Numerical answer [src]

7.27864599832406e-29

7.27864599832406e-29

The graph