What is the limit as t approaches 0+of arctan(1/t) ?

The limit as t approaches 0+of arctan(1/t) is pi/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

limit as t approaches 0+of arctan(1/t)

t \to 0 + lim (arctan (\frac{1}{t}))

\frac{π}{2}

Decimal

1.57079 \dots

Solution steps

t \to 0 + lim (arctan (\frac{1}{t}))

Apply the Limit Chain Rule:

\frac{π}{2}

= \frac{π}{2}

\int (4 cos (θ))^{2} d θ

\frac{\partial}{\partial t} (e^{s i n (x + c t)})

y^{^{''}} + 4 y^{^{'}} + 4 y = x^{- \frac{1}{2}}

\int 2 cos (x) sin (x) d x

\frac{d}{d x} (14.8 sin (\frac{π}{6} \cdot x - \frac{π}{2}) + 10)

What is the limit as t approaches 0+of arctan(1/t) ?
The limit as t approaches 0+of arctan(1/t) is pi/2