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 \infty lim (0.7 5^{x})

x \to 0 lim ((1 - 2 x) \frac{1}{x})

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

x \to 2 - lim (\frac{3}{( x - 2 )})

x \to 0 - lim (\frac{2}{x} - \frac{3}{x ^{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