Let f(x)=4+e1x1+e4x+2sinx|x|, then limx→0f(x)=

limx→0+f(x)=limx→0+4+e1x1+e4x+2sinxx=limx→0+4e−1x+1e−1x+e3x+2sinxx=0+2=2limx→0−f(x)=limx→0−4+e1x1+e4x−2sinxx=4−2=2Since the both limits are equal, the limit value exists and equal to 2.

Let f(x)=4+e1x1+e4x+2sinx|x|, then limx→0f(x)=

limx→0+f(x)=limx→0+4+e1x1+e4x+2sinxx=limx→0+4e−1x+1e−1x+e3x+2sinxx=0+2=2limx→0−f(x)=limx→0−4+e1x1+e4x−2sinxx=4−2=2Since the both limits are equal, the limit value exists and equal to 2.

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$Let f (x) = \frac{4 + e^{\frac{1}{x}}}{1 + e^{\frac{4}{x}}} + 2 \frac{\sin x}{| x |}, then \lim_{x \to 0} f (x) =$

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Detailed Solution

$\begin{matrix} \lim_{x \to 0 +} f (x) = \lim_{x \to 0 +} \frac{4 + e^{\frac{1}{x}}}{1 + e^{\frac{4}{x}}} + 2 \frac{\sin x}{x} \\ = \lim_{x \to 0 +} \frac{4 e^{\frac{- 1}{x}} + 1}{e^{\frac{- 1}{x}} + e^{\frac{3}{x}}} + 2 \frac{\sin x}{x} \\ = 0 + 2 \\ = 2 \end{matrix}$

$\begin{matrix} \lim_{x \to 0 -} f (x) = \lim_{x \to 0 -} \frac{4 + e^{\frac{1}{x}}}{1 + e^{\frac{4}{x}}} - 2 \frac{\sin x}{x} \\ = 4 - 2 \\ = 2 \end{matrix}$