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Limit of the function 1/(4+x^2)

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        1   
 lim  ------
x->-oo     2
      4 + x

\lim_{x \to -\infty} \frac{1}{x^{2} + 4}

Limit(1/(4 + x^2), x, -oo)

Detail solution

Let's take the limit

\lim_{x \to -\infty} \frac{1}{x^{2} + 4}

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

\lim_{x \to -\infty} \frac{1}{x^{2} + 4}

\lim_{x \to -\infty}\left(\frac{1}{x^{2} \left(1 + \frac{4}{x^{2}}\right)}\right)

Do Replacement

u = \frac{1}{x}

then

\lim_{x \to -\infty}\left(\frac{1}{x^{2} \left(1 + \frac{4}{x^{2}}\right)}\right) = \lim_{u \to 0^+}\left(\frac{u^{2}}{4 u^{2} + 1}\right)

\frac{0^{2}}{4 \cdot 0^{2} + 1} = 0

The final answer:

\lim_{x \to -\infty} \frac{1}{x^{2} + 4} = 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} \frac{1}{x^{2} + 4} = 0

\lim_{x \to \infty} \frac{1}{x^{2} + 4} = 0

\lim_{x \to 0^-} \frac{1}{x^{2} + 4} = \frac{1}{4}

\lim_{x \to 0^+} \frac{1}{x^{2} + 4} = \frac{1}{4}

\lim_{x \to 1^-} \frac{1}{x^{2} + 4} = \frac{1}{5}

\lim_{x \to 1^+} \frac{1}{x^{2} + 4} = \frac{1}{5}

The graph