What is the limit as h approaches 2 of 1/(4-h^2) ?

The limit as h approaches 2 of 1/(4-h^2) is diverges

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limit as h approaches 2 of 1/(4-h^2)

h \to 2 lim (\frac{1}{4 - h ^{2}})

diverges

Solution steps

h \to 2 lim (\frac{1}{4 - h ^{2}})

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

then the limit does not exist

h \to 2 + lim (\frac{1}{4 - h ^{2}}) = - \infty

h \to 2 - lim (\frac{1}{4 - h ^{2}}) = \infty

= diverges

x \to \infty lim (2 + 3 x)

x \to \infty lim (2^{x} - x)

x \to - 3 + lim (- \frac{x ^{2}}{3 x + 9})

x \to - 3 + lim (9 - x^{2})

h \to 0 lim (\frac{\frac{2 x + 2 h + 1}{2 - 3 x - 3 h} - \frac{2 x + 1}{2 - 3 x}}{h})

What is the limit as h approaches 2 of 1/(4-h^2) ?
The limit as h approaches 2 of 1/(4-h^2) is diverges