Function Continuity Condition
A function $f(x)$ is continuous at $x=a$ if $ \lim_{x \to a} f(x) = f(a) $. For the given function $f(x)$, we need $f(0) = \lim_{x \to 0} f(x)$.
Calculating the Limit of f(x) at x = 0
The function is $ f(x) = \frac{e^x\left(e^{\tan x - x} - 1\right) + \log_e(\sec x + \tan x) - x}{\tan x - x} $. Direct substitution yields the indeterminate form $ \frac{0}{0} $. We evaluate the limit using standard calculus methods.
Method: Limit Decomposition
Split the function into two parts:
- $ \text{Part 1:} \lim_{x \to 0} \frac{e^x\left(e^{\tan x - x} - 1\right)}{\tan x - x} $
- $ \text{Part 2:} \lim_{x \to 0} \frac{\log_e(\sec x + \tan x) - x}{\tan x - x} $
Evaluating Part 1
Let $u = \tan x - x$. As $x \to 0$, $u \to 0$. The limit becomes $ \lim_{x \to 0} e^x \cdot \frac{e^u - 1}{u} $. Using the standard limit $ \lim_{u \to 0} \frac{e^u - 1}{u} = 1 $, Part 1 evaluates to $ e^0 \cdot 1 = 1 $.
Evaluating Part 2
Evaluating $ \lim_{x \to 0} \frac{\log_e(\sec x + \tan x) - x}{\tan x - x} $ using limit properties (such as L'Hôpital's Rule or Taylor series expansions) yields a value of 1.
Combining the Results
The total limit is the sum of the limits of the two parts:
$ \lim_{x \to 0} f(x) = 1 + 1 = 2 $
Thus, for the function to be continuous at $x=0$, $f(0)$ must be equal to this limit.
$ f(0) = 2 $
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Important Questions from Differential Calculus
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