What is the limit as x approaches 0 of (e^{4x}-1)/(2x^2+2x) ?

The limit as x approaches 0 of (e^{4x}-1)/(2x^2+2x) is 2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

limit as x approaches 0 of (e^{4x}-1)/(2x^2+2x)

x \to 0 lim (\frac{e ^{4 x} - 1}{2 x ^{2} + 2 x})

2

Solution steps

x \to 0 lim (\frac{e ^{4 x} - 1}{2 x ^{2} + 2 x})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{4 e ^{4 x}}{4 x + 2})

= x \to 0 lim (\frac{4 e ^{4 x}}{4 x + 2})

Plug in the value

x = 0

= \frac{4 e ^{4 \cdot 0}}{4 \cdot 0 + 2}

Simplify

\frac{4 e ^{4 \cdot 0}}{4 \cdot 0 + 2} : 2

= 2

\int sin (x) sin (n) x d x

n = 1 \sum \infty \frac{4}{n ( n + 6 )}

\int sin^{4} (x) cos^{3} (x) d x

\int \frac{y ^{2} + 1}{( y - 1 ) ^{2} ( y + 1 )} d y

y^{^{''}} + 2 y^{^{'}} + 4 y = e^{t} cos (t)

What is the limit as x approaches 0 of (e^{4x}-1)/(2x^2+2x) ?
The limit as x approaches 0 of (e^{4x}-1)/(2x^2+2x) is 2