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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

limit as h approaches 0 of 1/(h^2)

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

\infty

Solution steps

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

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

then

x \to a lim f (x) = L

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

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

= \infty

\int x n (x) d x

\int sec^{13} (x) d x

f (x) = sinh (2 x)

\int_{0}^{l n (2)} e^{x + 2 y} d x

\frac{\partial}{\partial x} (2 x \cdot e^{x y})

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