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:

This limit can also be solved using L'Hôpital's Rule because substituting $x = \frac{\pi}{2}$ yields a $\frac{0}{0}$ indeterminate form. Differentiating the numerator gives $-\sin x$ and the denominator gives $-1$. Evaluating $\frac{-\sin(\pi/2)}{-1} = 1$ confirms the answer instantly!

Answer 1

Concept: For a function $f(x)$ to be continuous at a specific point $x = c$, three conditions must be satisfied: the function must be defined at $c$, the limit as $x \to c$ must exist, and the value of the limit must exactly equal the functional value at that point.
• Continuity Condition: $\lim_{x \to c} f(x) = f(c)$
• Standard Limit Identity: $\lim_{h \to 0} \frac{\sin h}{h} = 1$

Step 1: Evaluate the limit of $f(x)$ as $x$ approaches $\frac{\pi}{2}$ using variable substitution.
We are given $f\left(\frac{\pi}{2}\right) = 1$. Let us compute the limit: \[ L = \lim_{x \to \frac{\pi}{2}} f(x) = \lim_{x \to \frac{\pi}{2}} \frac{\cos x}{\frac{\pi}{2} - x} \] To evaluate this limit, let us make a change of variable. Let $\frac{\pi}{2} - x = h$. As $x \to \frac{\pi}{2}$, it follows that $h \to 0$. Also, we can express $x$ as $x = \frac{\pi}{2} - h$. Substitute these into the limit expression: \[ L = \lim_{h \to 0} \frac{\cos\left(\frac{\pi}{2} - h\right)}{h} \] Using the trigonometric co-function identity $\cos\left(\frac{\pi}{2} - h\right) = \sin h$: \[ L = \lim_{h \to 0} \frac{\sin h}{h} \]

Step 2: Apply standard limit laws and conclude the proof.
Using the standard fundamental trigonometric limit theorem, we know: \[ \lim_{h \to 0} \frac{\sin h}{h} = 1 \] Therefore, we find that: \[ \lim_{x \to \frac{\pi}{2}} f(x) = 1 \] We are already given that the value of the function at the point is: \[ f\left(\frac{\pi}{2}\right) = 1 \] Since $\lim_{x \to \frac{\pi}{2}} f(x) = f\left(\frac{\pi}{2}\right) = 1$, all conditions for continuity are fulfilled. Thus, the function $f(x)$ is continuous at $x = \frac{\pi}{2}$.