Question

$\lim_{x \to 0} \frac{e^{\tan x} - e^x}{\tan x - x} =$}

• A. $1$
• B. $0$
• C. $\frac{1}{2}$
• D. $\frac{1}{4}$

Hint:

If you see $e^A - e^B$, always factor out $e^B$ to create the form $e^{A-B} - 1$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This is a limit problem of the form $\frac{0}{0}$. We can manipulate the expression algebraically to use standard limits.
Step 2: Key Formula or Approach:
Standard limit: $\lim_{u \to 0} \frac{e^u - 1}{u} = 1$.
We will factor out $e^x$ from the numerator to create an expression that matches this standard form.
Step 3: Detailed Explanation:
We need to evaluate:
\[ L = \lim_{x \to 0} \frac{e^{\tan x} - e^x}{\tan x - x} \]
Factor out $e^x$ from the numerator:
\[ L = \lim_{x \to 0} \frac{e^x(e^{\tan x} \cdot e^{-x} - 1)}{\tan x - x} \]
\[ L = \lim_{x \to 0} \frac{e^x(e^{\tan x - x} - 1)}{\tan x - x} \]
We can use the properties of limits to separate this into a product of limits:
\[ L = \left( \lim_{x \to 0} e^x \right) \cdot \left( \lim_{x \to 0} \frac{e^{\tan x - x} - 1}{\tan x - x} \right) \]
Let $u = \tan x - x$.
As $x \to 0$, we know $\tan x \to 0$, so $u \to 0 - 0 = 0$.
The second limit becomes the standard exponential limit:
\[ \lim_{u \to 0} \frac{e^u - 1}{u} = 1 \]
The first limit is simply evaluating the continuous function $e^x$ at 0:
\[ \lim_{x \to 0} e^x = e^0 = 1 \]
Multiplying the two results:
\[ L = 1 \cdot 1 = 1 \]
Step 4: Final Answer:
The limit evaluates to $1$.