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

The limit as x approaches 0 of (e^x-cos(x))/(ln(1+x)) is 1

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limit as x approaches 0 of (e^x-cos(x))/(ln(1+x))

x \to 0 lim (\frac{e ^{x} - cos ( x )}{ln ( 1 + x )})

1

Solution steps

x \to 0 lim (\frac{e ^{x} - cos ( x )}{ln ( 1 + x )})

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

= \frac{e ^{0} + sin ( 0 )}{\frac{1}{1 + 0}}

Simplify

\frac{e ^{0} + sin ( 0 )}{\frac{1}{1 + 0}} : 1

= 1

x \to 0 lim (\frac{7 x - 1}{5 x})

x \to 1 lim (2 x - 4)

x \to 0 lim (\frac{x ^{2} - x}{∣ x ∣})

x \to 0 + lim (\frac{1}{tan ( x )})

x \to 4 - lim (x - 2)

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