Discussion
Consider the function f defined by :
\[ f(x) = \begin{cases} \dfrac{\lambda \cos(x)}{\pi/2 - x}, & x \neq \pi/2 \\ 1, & x = \pi/2 \end{cases} \]
What should be the value of \(\lambda\) so that f would be continuous at \(x = \pi/2\) ?
Answer: A
The value is \(\lambda = 1\).
Continuity at \(\pi/2\) needs the limit of \(f\) to equal \(f(\pi/2) = 1\). Put \(h = \pi/2 - x\); then \(\cos x = \cos(\pi/2 - h) = \sin h\), and
\[ \begin{aligned} \lim_{x \to \pi/2} \frac{\lambda \cos x}{\pi/2 - x} &= \lim_{h \to 0} \frac{\lambda \sin h}{h} \\ &= \lambda \end{aligned} \]
Setting \(\lambda = 1\) makes the limit equal the function value. A sign slip (writing \(x - \pi/2\) in the denominator) would give \(-1\), and the values 2 and \(-2\) come from no correct step.
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