What is the limit as t approaches 0 of (sqrt(t^2+9-3))/(t^2) ?

The limit as t approaches 0 of (sqrt(t^2+9-3))/(t^2) is infinity

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limit as t approaches 0 of (sqrt(t^2+9-3))/(t^2)

t \to 0 lim (\frac{t ^{2} + 9 - 3}{t ^{2}})

\infty

Solution steps

t \to 0 lim (\frac{t ^{2} + 9 - 3}{t ^{2}})

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

then

x \to a lim f (x) = L

t \to 0 - lim (\frac{t ^{2} + 9 - 3}{t ^{2}}) = \infty

t \to 0 + lim (\frac{t ^{2} + 9 - 3}{t ^{2}}) = \infty

= \infty

\int \frac{1}{2 x - 1} - \frac{1}{2 x - 1} d x

x y^{^{'}} = \frac{x ^{3} cos ( 2 x )}{y} + y

\frac{d}{d x} (\frac{x - 5}{8 x})

y = \frac{( x + 8 )}{( x - 4 )}

\frac{d}{d x} (a + b x + c x^{2})

What is the limit as t approaches 0 of (sqrt(t^2+9-3))/(t^2) ?
The limit as t approaches 0 of (sqrt(t^2+9-3))/(t^2) is infinity