What is the limit as x approaches 0 of 1/(4x^2) ?

The limit as x approaches 0 of 1/(4x^2) is infinity

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limit as x approaches 0 of 1/(4x^2)

x \to 0 lim (\frac{1}{4 x ^{2}})

\infty

Solution steps

x \to 0 lim (\frac{1}{4 x ^{2}})

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

then

x \to a lim f (x) = L

x \to 0 - lim (\frac{1}{4 x ^{2}}) = \infty

x \to 0 + lim (\frac{1}{4 x ^{2}}) = \infty

= \infty

f (x) = 1.8 sin (x) - x

\int \frac{15}{( x - 1 ) ( x ^{2} + 4 )} d x

f (x) = e^{5 x}

\frac{d y}{d x} = \frac{40}{x ^{- 5}}

k = 5 \sum \infty \frac{1}{( k - 2 ) ^{4}}

What is the limit as x approaches 0 of 1/(4x^2) ?
The limit as x approaches 0 of 1/(4x^2) is infinity