What is the limit as x approaches 0+of 3e^x+cos(2x) ?

The limit as x approaches 0+of 3e^x+cos(2x) is 4

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limit as x approaches 0+of 3e^x+cos(2x)

x \to 0 + lim (3 e^{x} + cos (2 x))

4

Solution steps

x \to 0 + lim (3 e^{x} + cos (2 x))

Plug in the value

x = 0

= 3 e^{0} + cos (2 \cdot 0)

Simplify

3 e^{0} + cos (2 \cdot 0) : 4

= 4

x \to 0 lim (\frac{x + 12}{x + 6})

x \to 3 + lim ((3 - x) \cdot ln (x - 3))

x \to 0 lim (\frac{tan ( 9 x )}{sin ( 6 x )})

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

x \to - 1 lim (- x^{5} + 6 x^{4} + 2)

What is the limit as x approaches 0+of 3e^x+cos(2x) ?
The limit as x approaches 0+of 3e^x+cos(2x) is 4