Limit of the function 1/(8+x)

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        1  
 lim  -----
x->-8+8 + x

\lim_{x \to -8^+} \frac{1}{x + 8}

Limit(1/(8 + x), x, -8)

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

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Rapid solution [src]

oo

\infty

Expand and simplify

One‐sided limits [src]

        1  
 lim  -----
x->-8+8 + x

\lim_{x \to -8^+} \frac{1}{x + 8}

oo

\infty

= 151.0

        1  
 lim  -----
x->-8-8 + x

\lim_{x \to -8^-} \frac{1}{x + 8}

-oo

-\infty

= -151.0

= -151.0

Other limits x→0, -oo, +oo, 1

\lim_{x \to -8^-} \frac{1}{x + 8} = \infty

\lim_{x \to -8^+} \frac{1}{x + 8} = \infty

\lim_{x \to \infty} \frac{1}{x + 8} = 0

\lim_{x \to 0^-} \frac{1}{x + 8} = \frac{1}{8}

\lim_{x \to 0^+} \frac{1}{x + 8} = \frac{1}{8}

\lim_{x \to 1^-} \frac{1}{x + 8} = \frac{1}{9}

\lim_{x \to 1^+} \frac{1}{x + 8} = \frac{1}{9}

\lim_{x \to -\infty} \frac{1}{x + 8} = 0

Numerical answer [src]

151.0

151.0

The graph