11) $lim_{x \to 0}\frac{1 - sin(\frac{pi}{2}-x)}{x}$

Answer

Explanation:

Step1: Simplify the trigonometric term.

Use the co-function identity $\sin\left(\frac{\pi}{2} - x\right) = \cos(x)$. $$ \lim_{x \to 0} \frac{1 - \sin\left(\frac{\pi}{2} - x\right)}{x} = \lim_{x \to 0} \frac{1 - \cos(x)}{x} $$

Step2: Check for indeterminate form.

Substituting $x=0$ into the simplified expression gives $\frac{1 - \cos(0)}{0} = \frac{1-1}{0} = \frac{0}{0}$.

Step3: Apply L'Hopital's Rule.

Since the limit is in the indeterminate form $\frac{0}{0}$, we differentiate the numerator and the denominator with respect to $x$. $$ \lim_{x \to 0} \frac{\frac{d}{dx}(1 - \cos(x))}{\frac{d}{dx}(x)} = \lim_{x \to 0} \frac{-(-\sin(x))}{1} = \lim_{x \to 0} \frac{\sin(x)}{1} $$

Step4: Evaluate the resulting limit.

Substitute $x=0$ into the expression from Step 3. $$ \frac{\sin(0)}{1} = \frac{0}{1} = 0 $$