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Limit of the function 8/(16-x^2)

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      /   8   \
 lim  |-------|
x->-oo|      2|
      \16 - x /

\lim_{x \to -\infty}\left(\frac{8}{16 - x^{2}}\right)

Limit(8/(16 - x^2), x, -oo)

Detail solution

Let's take the limit

\lim_{x \to -\infty}\left(\frac{8}{16 - x^{2}}\right)

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

\lim_{x \to -\infty}\left(\frac{8}{16 - x^{2}}\right)

\lim_{x \to -\infty}\left(\frac{8 \frac{1}{x^{2}}}{-1 + \frac{16}{x^{2}}}\right)

Do Replacement

u = \frac{1}{x}

then

\lim_{x \to -\infty}\left(\frac{8 \frac{1}{x^{2}}}{-1 + \frac{16}{x^{2}}}\right) = \lim_{u \to 0^+}\left(\frac{8 u^{2}}{16 u^{2} - 1}\right)

\frac{8 \cdot 0^{2}}{-1 + 16 \cdot 0^{2}} = 0

The final answer:

\lim_{x \to -\infty}\left(\frac{8}{16 - x^{2}}\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

Rapid solution [src]

0

Expand and simplify

Other limits x→0, -oo, +oo, 1

\lim_{x \to -\infty}\left(\frac{8}{16 - x^{2}}\right) = 0

\lim_{x \to \infty}\left(\frac{8}{16 - x^{2}}\right) = 0

\lim_{x \to 0^-}\left(\frac{8}{16 - x^{2}}\right) = \frac{1}{2}

\lim_{x \to 0^+}\left(\frac{8}{16 - x^{2}}\right) = \frac{1}{2}

\lim_{x \to 1^-}\left(\frac{8}{16 - x^{2}}\right) = \frac{8}{15}

\lim_{x \to 1^+}\left(\frac{8}{16 - x^{2}}\right) = \frac{8}{15}

The graph