Key Concepts and Formulas

• Limit of the form $1^\infty$: If a limit is of the indeterminate form $1^\infty$, i.e., $\mathop {\lim }\limits_{x \to a} f(x) = 1$ and $\mathop {\lim }\limits_{x \to a} g(x) = \infty$, then $\mathop {\lim }\limits_{x \to a} [f(x)]^{g(x)} = e^{\mathop {\lim }\limits_{x \to a} [f(x) - 1]g(x)}$.
• Trigonometric Identities:
  • $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$
  • $\tan(\frac{\pi}{4}) = 1$
• Standard Limits:
  • $\mathop {\lim }\limits_{x \to 0} \frac{\tan x}{x} = 1$

Step-by-Step Solution

Step 1: Identify the Indeterminate Form We are asked to evaluate the limit: $L = \mathop {\lim }\limits_{x \to 0} {\left( {\tan \left( {{\pi \over 4} + x} \right)} \right)^{{1 \over x}}}$ As $x \to 0$, we have: $\tan \left( {{\pi \over 4} + x} \right) \to \tan \left( {{\pi \over 4}} \right) = 1$. And the exponent $\frac{1}{x} \to \frac{1}{0}$, which tends to $\infty$ (or $-\infty$, depending on the direction of approach, but for the form $1^\infty$, we consider the magnitude). Thus, the limit is of the indeterminate form $1^\infty$.

Step 2: Apply the $1^\infty$ Limit Formula For a limit of the form $1^\infty$, we use the formula $e^{\mathop {\lim }\limits_{x \to a} [f(x) - 1]g(x)}$. Here, $f(x) = \tan \left( {{\pi \over 4} + x} \right)$ and $g(x) = \frac{1}{x}$. So, the limit becomes: $L = e^{\mathop {\lim }\limits_{x \to 0} \left[ {\tan \left( {{\pi \over 4} + x} \right) - 1} \right] \times \frac{1}{x}}$

Step 3: Simplify the Trigonometric Term We use the tangent addition formula $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ with $A = \frac{\pi}{4}$ and $B = x$. $\tan \left( {{\pi \over 4} + x} \right) = \frac{\tan \left( {{\pi \over 4}} \right) + \tan x}{1 - \tan \left( {{\pi \over 4}} \right) \tan x}$ Since $\tan \left( {{\pi \over 4}} \right) = 1$, we get: $\tan \left( {{\pi \over 4} + x} \right) = \frac{1 + \tan x}{1 - \tan x}$ Now, substitute this back into the expression inside the exponent: ${\tan \left( {{\pi \over 4} + x} \right) - 1} = \frac{1 + \tan x}{1 - \tan x} - 1$ To simplify, find a common denominator: $= \frac{1 + \tan x - (1 - \tan x)}{1 - \tan x} = \frac{1 + \tan x - 1 + \tan x}{1 - \tan x} = \frac{2 \tan x}{1 - \tan x}$

Step 4: Substitute the Simplified Term and Evaluate the Limit in the Exponent Now, substitute this simplified expression back into the exponent: $L = e^{\mathop {\lim }\limits_{x \to 0} \left[ \frac{2 \tan x}{1 - \tan x} \right] \times \frac{1}{x}}$ $L = e^{\mathop {\lim }\limits_{x \to 0} \frac{2 \tan x}{x(1 - \tan x)}}$ We can rewrite this as: $L = e^{2 \mathop {\lim }\limits_{x \to 0} \left[ \frac{\tan x}{x} \times \frac{1}{1 - \tan x} \right]}$

Step 5: Evaluate the Limit using Standard Limits and Substitution We know that $\mathop {\lim }\limits_{x \to 0} \frac{\tan x}{x} = 1$. Also, as $x \to 0$, $\tan x \to \tan 0 = 0$. So, the limit in the exponent becomes: $2 \mathop {\lim }\limits_{x \to 0} \left[ \frac{\tan x}{x} \right] \times \mathop {\lim }\limits_{x \to 0} \left[ \frac{1}{1 - \tan x} \right]$ $= 2 \times 1 \times \frac{1}{1 - 0}$ $= 2 \times 1 \times 1 = 2$

Step 6: State the Final Result Substituting this value back into the expression for $L$: $L = e^2$

Common Mistakes & Tips

• Incorrectly identifying the indeterminate form: Always check the form of the limit before applying any formulas. If it's not $1^\infty$, the $e^{\lim (f-1)g}$ formula cannot be directly applied.
• Algebraic errors in trigonometric simplification: Be very careful when expanding $\tan(\frac{\pi}{4} + x)$ and simplifying the expression $\tan(\frac{\pi}{4} + x) - 1$. Small mistakes here will lead to an incorrect final answer.
• Forgetting the standard limit $\mathop {\lim }\limits_{x \to 0} \frac{\tan x}{x} = 1$: This limit is crucial for simplifying the expression in the exponent.

Summary

The given limit is of the indeterminate form $1^\infty$. We first applied the standard formula for such limits, which transforms the problem into evaluating $e$ raised to a new limit. We then used the tangent addition formula to simplify the base of the original expression and subsequently simplified the expression within the exponent. Finally, by utilizing the standard limit $\mathop {\lim }\limits_{x \to 0} \frac{\tan x}{x} = 1$, we evaluated the exponent to be 2, leading to the final answer of $e^2$.

The final answer is $\boxed{e^2}$.