What is the limit as x approaches 0+of (e^x-1)cot(x) ?

The limit as x approaches 0+of (e^x-1)cot(x) is 1

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limit as x approaches 0+of (e^x-1)cot(x)

x \to 0 + lim ((e^{x} - 1) cot (x))

1

Solution steps

x \to 0 + lim ((e^{x} - 1) cot (x))

Rewrite for L'Hopital

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

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

= \frac{e ^{0}}{sec ^{2} ( 0 )}

Simplify

\frac{e ^{0}}{sec ^{2} ( 0 )} : 1

= 1

x \to 0 lim (\frac{ln ( x ^{2} + 1 )}{x + 4 \cdot sin ( x )})

x \to 0 lim (x^{3} \cdot cos (\frac{1}{x}))

x \to 0 lim (\frac{e ^{x}}{1})

n \to \infty lim (n)

x \to \infty lim (- 1^{x} \cdot x)

What is the limit as x approaches 0+of (e^x-1)cot(x) ?
The limit as x approaches 0+of (e^x-1)cot(x) is 1