Determine \( \lim _{x \rightarrow \frac{\pi^{+}}{}+} e^{\tan x} \)

To find the limit \( \lim_{x \rightarrow \pi^+} e^{\tan x} \), we need to first analyze the behavior of \( \tan x \) as \( x \) approaches \( \pi \) from the right. The tangent function, \( \tan x \), is defined as \( \frac{\sin x}{\cos x} \). As \( x \) approaches \( \pi \): - \( \sin(\pi) = 0 \) - \( \cos(\pi) = -1 \) Thus, \( \tan(\pi) = \frac{\sin(\pi)}{\cos(\pi)} = \frac{0}{-1} = 0 \). Now, we consider the behavior of \( \tan x \) for values of \( x \) slightly greater than \( \pi \). When \( x \) is slightly more than \( \pi \), the sine function remains near zero, while the cosine function goes slightly negative (since it changes from -1, dropping slightly down as we pass π). This means that as \( x \rightarrow \pi^+ \), \( \tan x \) approaches 0 from the negative side: \[ \tan x \to 0^- \quad \text{as } x \rightarrow \pi^+ \] Now we can substitute this behavior into our limit of the exponential function: \[ \lim_{x \rightarrow \pi^+} \tan x = 0^- \] Therefore, as \( x \) approaches \( \pi \) from the right, \( e^{\tan x} \) will be closer to \( e^{0^-} \), which is practically the same as \( e^0 \): \[ e^{\tan x} \to e^0 = 1 \] Thus, we conclude that: \[ \lim_{x \rightarrow \pi^+} e^{\tan x} = 1 \] In summary, the solution to the limit is: \[ \boxed{1} \]