The value of `lim_(xrarr0)(sin^(2)x+cosx-1)/(x^(2))` is

Options

A. 0
B. `(1)/(2)`
C. `-(1)/(2)`
D. 0

Answer

To solve the limit \( \lim_{x \to 0} \frac{\sin^2 x + \cos x - 1}{x^2} \), we can break it down step by step.

### Step 1: Rewrite the limit
We start with the limit:
\[ L = \lim_{x \to 0} \frac{\sin^2 x + \cos x - 1}{x^2} \]
We can separate the limit into two parts:
\[ L = \lim_{x \to 0} \frac{\sin^2 x}{x^2} + \lim_{x \to 0} \frac{\cos x - 1}{x^2} \]

### Step 2: Evaluate the first limit
The first limit is:
\[ \lim_{x \to 0} \frac{\sin^2 x}{x^2} \]
Using the standard limit \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \), we have:
\[ \lim_{x \to 0} \frac{\sin^2 x}{x^2} = \left(\lim_{x \to 0} \frac{\sin x}{x}\right)^2 = 1^2 = 1 \]

### Step 3: Evaluate the second limit
Now we evaluate the second limit:
\[ \lim_{x \to 0} \frac{\cos x - 1}{x^2} \]
Using the Taylor series expansion for \( \cos x \) around \( x = 0 \):
\[ \cos x \approx 1 - \frac{x^2}{2} + O(x^4) \]
Thus,
\[ \cos x - 1 \approx -\frac{x^2}{2} \]
So,
\[ \lim_{x \to 0} \frac{\cos x - 1}{x^2} = \lim_{x \to 0} \frac{-\frac{x^2}{2}}{x^2} = -\frac{1}{2} \]

### Step 4: Combine the results
Now we combine the results from Step 2 and Step 3:
\[ L = 1 - \frac{1}{2} = \frac{1}{2} \]

### Final Answer
Thus, the value of the limit is:
\[ \boxed{\frac{1}{2}} \]