What is the limit as x approaches (pi/2)+of 9e^{tan(x)} ?

The limit as x approaches (pi/2)+of 9e^{tan(x)} is 0

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

limit as x approaches (pi/2)+of 9e^{tan(x)}

x \to (\frac{π}{2}) + lim (9 e^{t a n (x)})

0

Solution steps

x \to \frac{π}{2} + lim (9 e^{t a n (x)})

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

= 9 \cdot x \to \frac{π}{2} + lim (e^{t a n (x)})

Apply the Limit Chain Rule:

0

= 9 \cdot 0

Simplify

= 0

\frac{d}{d x} (5 x^{4} + x^{2})

\int \frac{y ^{2}}{( 49 - y ^{2} ) ^{\frac{3}{2}}} d y

\int (3 x - 6) e^{x^{2} - 4 x} d x

cos^{2} (x) sin^{2} (x)

n = 1 \sum \infty \frac{n !}{10 5 ^{n}}

What is the limit as x approaches (pi/2)+of 9e^{tan(x)} ?
The limit as x approaches (pi/2)+of 9e^{tan(x)} is 0