Question

Show that the function \( f(x) = \begin{cases \frac{\cos x}{\frac{\pi}{2}-x}, & x \neq \frac{\pi}{2} \\ 1, & x = \frac{\pi}{2} \end{cases} \) is continuous at \( x = \frac{\pi}{2} \).}

Hint:

For limits involving \( \frac{\pi}{2} - x \), substituting \( h = \frac{\pi}{2} - x \) often converts the problem into a standard limit involving \( \sin h / h \).
Always state the condition for continuity explicitly in descriptive answers.

Answer 1

Concept:
• A function \( f(x) \) is continuous at a point \( x = a \) if \( \lim_{x \to a} f(x) = f(a) \).
• This requires the limit to exist and be equal to the functional value at that point.
• We can use the substitution method or L'Hôpital's rule to evaluate indeterminate forms like \( 0/0 \).

Step 1: Determine the functional value at the point
From the definition of the function given in the problem:
\( f\left(\frac{\pi}{2}\right) = 1 \).

Step 2: Evaluate the limit as \( x \) approaches \( \frac{\pi}{2} \)
The limit to evaluate is:
\[ L = \lim_{x \to \frac{\pi}{2}} \frac{\cos x}{\frac{\pi}{2} - x} \] As \( x \to \frac{\pi}{2} \), the expression is in the indeterminate form \( \frac{\cos(\pi/2)}{\pi/2 - \pi/2} = \frac{0}{0} \).

Step 3: Apply substitution to simplify the limit
Let \( \frac{\pi}{2} - x = h \). As \( x \to \frac{\pi}{2} \), \( h \to 0 \).
Also, \( x = \frac{\pi}{2} - h \).
Substituting these into the limit expression:
\[ L = \lim_{h \to 0} \frac{\cos\left(\frac{\pi}{2} - h\right)}{h} \] Using the trigonometric identity \( \cos\left(\frac{\pi}{2} - \theta\right) = \sin \theta \):
\[ L = \lim_{h \to 0} \frac{\sin h}{h} \]

Step 4: Compare the limit with the functional value
We know the standard limit \( \lim_{h \to 0} \frac{\sin h}{h} = 1 \).
So, \( \lim_{x \to \frac{\pi}{2}} f(x) = 1 \).
Since \( \lim_{x \to \frac{\pi}{2}} f(x) = f\left(\frac{\pi}{2}\right) = 1 \), the function is continuous at \( x = \frac{\pi}{2} \).