The value of lim_(x→0) (1 - cosx)/x² is:

The correct option is 1/2.

Step 1: Identify the Limit and Check for Indeterminate Form
We are asked to evaluate the limit:
$$\underset{x\to 0}{lim}\frac{1-cosx}{x^2}$$
Let us check the values of the numerator and denominator as $x$ approaches $0$:
• For the numerator: $1-cos\left(0\right)=1-1=0$
• For the denominator: $0^2=0$
As shown in the provided image, substituting $x=0$ directly yields the indeterminate form:
$$\frac{0}{0}$$

Step 2: Apply L'Hospital's Rule
Since the limit yields a $\frac{0}{0}$ form, we can apply L'Hospital's Rule (labeled as "Apply Hospital Rule" in the image). This rule states that we can differentiate the numerator and the denominator separately with respect to $x$:
• Differentiating the numerator:
$$\frac{d}{dx}(1-cosx)=sinx$$
• Differentiating the denominator:
$$\frac{d}{dx}\left(x^2\right)=2x$$
Substituting these derivatives back into the limit gives:
$$\underset{x\to 0}{lim}\frac{sinx}{2x}$$

Step 3: Simplify and Evaluate
We factor out the constant factor $\frac{1}{2}$ from the limit expression:
$$\frac{1}{2}\underset{x\to 0}{lim}\frac{sinx}{x}$$
Using the standard trigonometric limit identity:
$$\underset{x\to 0}{lim}\frac{sinx}{x}=1$$
We multiply the result by our factored coefficient:
$$\frac{1}{2}\times 1=\frac{1}{2}$$
Therefore, the limit is $\frac{1}{2}$.