What is the limit as t approaches 0+of 2/t-1/(e^t-1) ?

The limit as t approaches 0+of 2/t-1/(e^t-1) is infinity

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

limit as t approaches 0+of 2/t-1/(e^t-1)

t \to 0 + lim (\frac{2}{t} - \frac{1}{e ^{t} - 1})

\infty

Solution steps

t \to 0 + lim (\frac{2}{t} - \frac{1}{e ^{t} - 1})

\frac{2}{t} - \frac{1}{e ^{t} - 1} = \frac{2}{t} (1 - \frac{\frac{1}{e ^{t} - 1}}{\frac{2}{t}})

= t \to 0 + lim (\frac{2}{t} (1 - \frac{\frac{1}{e ^{t} - 1}}{\frac{2}{t}}))

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

= 2 \cdot t \to 0 + lim (\frac{1}{t} (1 - \frac{\frac{1}{e ^{t} - 1}}{\frac{2}{t}}))

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

With the exception of indeterminate form

= 2 \cdot t \to 0 + lim (\frac{1}{t}) \cdot t \to 0 + lim (1 - \frac{\frac{1}{e ^{t} - 1}}{\frac{2}{t}})

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

t \to 0 + lim (1 - \frac{\frac{1}{e ^{t} - 1}}{\frac{2}{t}}) = \frac{1}{2}

= 2 \cdot \infty \cdot \frac{1}{2}

Apply Infinity Property:

c \cdot \infty = \infty

= \infty

What is the limit as t approaches 0+of 2/t-1/(e^t-1) ?
The limit as t approaches 0+of 2/t-1/(e^t-1) is infinity