`lim_(x->0)(1/(x^2)-1/(tan^2x))`

Options

A. 0
B. `1/3`
C. `2/3`
D. `1/2`

Answer

To solve the limit \( \lim_{x \to 0} \left( \frac{1}{x^2} - \frac{1}{\tan^2 x} \right) \), we will follow these steps:

### Step 1: Rewrite the Limit
We start by rewriting the limit expression:
\[ \lim_{x \to 0} \left( \frac{1}{x^2} - \frac{1}{\tan^2 x} \right) = \lim_{x \to 0} \left( \frac{\tan^2 x - x^2}{x^2 \tan^2 x} \right) \]

### Step 2: Find a Common Denominator
To combine the fractions, we take the common denominator \( x^2 \tan^2 x \):
\[ = \lim_{x \to 0} \frac{\tan^2 x - x^2}{x^2 \tan^2 x} \]

### Step 3: Use the Identity for \(\tan^2 x\)
Recall that \( \tan x = \frac{\sin x}{\cos x} \), hence:
\[ \tan^2 x = \frac{\sin^2 x}{\cos^2 x} \]
Thus, we can rewrite the limit as:
\[ = \lim_{x \to 0} \frac{\frac{\sin^2 x}{\cos^2 x} - x^2}{x^2 \frac{\sin^2 x}{\cos^2 x}} \]

### Step 4: Simplify the Expression
Now, we simplify the numerator:
\[ = \lim_{x \to 0} \frac{\sin^2 x - x^2 \cos^2 x}{x^2 \sin^2 x} \]

### Step 5: Apply Taylor Series Expansion
Using the Taylor series expansion for \( \sin x \):
\[ \sin x \approx x - \frac{x^3}{6} + O(x^5) \]
Thus,
\[ \sin^2 x \approx \left(x - \frac{x^3}{6}\right)^2 = x^2 - \frac{x^4}{3} + O(x^6) \]
Substituting this into our limit gives:
\[ = \lim_{x \to 0} \frac{x^2 - \frac{x^4}{3} - x^2 \cos^2 x}{x^2 \sin^2 x} \]

### Step 6: Expand \(\cos^2 x\)
Using the Taylor series for \( \cos x \):
\[ \cos x \approx 1 - \frac{x^2}{2} + O(x^4) \]
Thus,
\[ \cos^2 x \approx 1 - x^2 + O(x^4) \]
Now substituting this into our limit:
\[ = \lim_{x \to 0} \frac{x^2 - \frac{x^4}{3} - x^2(1 - x^2)}{x^2 \left(x^2 - \frac{x^4}{3}\right)} \]

### Step 7: Simplify Further
This simplifies to:
\[ = \lim_{x \to 0} \frac{-\frac{x^4}{3}}{x^2 \left(x^2 - \frac{x^4}{3}\right)} \]

### Step 8: Cancel Common Terms
Now we can cancel \( x^2 \):
\[ = \lim_{x \to 0} \frac{-\frac{x^2}{3}}{x^2 - \frac{x^4}{3}} = \lim_{x \to 0} \frac{-\frac{1}{3}}{1 - \frac{x^2}{3}} = -\frac{1}{3} \]

### Step 9: Evaluate the Limit
As \( x \to 0 \), \( \frac{x^2}{3} \) approaches \( 0 \):
\[ = -\frac{1}{3} \]

### Final Answer
Thus, the limit is:
\[ \lim_{x \to 0} \left( \frac{1}{x^2} - \frac{1}{\tan^2 x} \right) = \frac{2}{3} \]