What is the limit as x approaches 2 of 5/(x^4-16) ?

The limit as x approaches 2 of 5/(x^4-16) is diverges

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limit as x approaches 2 of 5/(x^4-16)

x \to 2 lim (\frac{5}{x ^{4} - 16})

diverges

Solution steps

x \to 2 lim (\frac{5}{x ^{4} - 16})

x \to a - lim f (x) \neq = x \to a + lim f (x)

then the limit does not exist

x \to 2 + lim (\frac{5}{x ^{4} - 16}) = \infty

x \to 2 - lim (\frac{5}{x ^{4} - 16}) = - \infty

= diverges

t \to 0 lim ((\frac{3}{t 1 + t}) - \frac{3}{t})

\int 2 + 4 y^{2} d y

\frac{d}{d x} (x 49 - x^{2})

\int x^{2} + 5 d x

\int \frac{1}{8 - 6 x - 9 x ^{2}} d x

What is the limit as x approaches 2 of 5/(x^4-16) ?
The limit as x approaches 2 of 5/(x^4-16) is diverges