What is the limit as x approaches 0 of 1/x+1/(t^2+1) ?

The limit as x approaches 0 of 1/x+1/(t^2+1) is diverges

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limit as x approaches 0 of 1/x+1/(t^2+1)

x \to 0 lim (\frac{1}{x} + \frac{1}{t ^{2} + 1})

diverges

Solution steps

x \to 0 lim (\frac{1}{x} + \frac{1}{t ^{2} + 1})

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

then the limit does not exist

x \to 0 + lim (\frac{1}{x} + \frac{1}{t ^{2} + 1}) = \infty

x \to 0 - lim (\frac{1}{x} + \frac{1}{t ^{2} + 1}) = - \infty

= diverges

x \to 0 lim (\frac{( 4 sin ( \frac{π}{2} + x ) - 4 sin ( \frac{π}{2} ) )}{x})

x \to - 1 + lim (\frac{- 1}{x + 1})

x \to 2 lim ((x^{2} + 3) (x - 4))

x \to 1 - lim (\frac{- 3 ( x ^{2} - 1 )}{2 x ^{2} + 2 x - 4})

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

What is the limit as x approaches 0 of 1/x+1/(t^2+1) ?
The limit as x approaches 0 of 1/x+1/(t^2+1) is diverges