What is the value of `lim_(x to 0)x/(tanx)`?

Answer

To find the value of the limit \( \lim_{x \to 0} \frac{x}{\tan x} \), we can follow these steps:

### Step 1: Recognize the Limit Form
As \( x \) approaches \( 0 \), both the numerator \( x \) and the denominator \( \tan x \) approach \( 0 \). This gives us an indeterminate form of \( \frac{0}{0} \).

### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that if \( \lim_{x \to c} \frac{f(x)}{g(x)} \) results in \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then:
\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]
provided the limit on the right exists.

### Step 3: Differentiate the Numerator and Denominator
Let \( f(x) = x \) and \( g(x) = \tan x \).
- The derivative of \( f(x) = x \) is \( f'(x) = 1 \).
- The derivative of \( g(x) = \tan x \) is \( g'(x) = \sec^2 x \).

### Step 4: Substitute the Derivatives into the Limit
Now we can substitute the derivatives into the limit:
\[ \lim_{x \to 0} \frac{x}{\tan x} = \lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \frac{1}{\sec^2 x} \]

### Step 5: Evaluate the Limit
As \( x \) approaches \( 0 \), \( \sec^2 x \) approaches \( \sec^2(0) = 1 \). Therefore:
\[ \lim_{x \to 0} \frac{1}{\sec^2 x} = \frac{1}{1} = 1 \]

### Final Result
Thus, the value of the limit is:
\[ \lim_{x \to 0} \frac{x}{\tan x} = 1 \]

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