Question

3 mark for review
\\( \lim _ { x \
ightarrow \pi } \frac { x + \pi \sec x } { x ^ { 2 } - \pi ^ { 2 } } \\) is
a \\( - \frac { \pi } { 2 } \\)
b 0
c \\( \frac { 1 } { 2 \pi } \\)
d nonexistent

Explanation:

Step1: Check indeterminate form

When \(x = \pi\), numerator \(x+\pi\sec x=\pi+\pi\sec\pi=\pi-\pi = 0\), denominator \(x^{2}-\pi^{2}=\pi^{2}-\pi^{2}=0\). So, it's \(\frac{0}{0}\) form, apply L'Hopital's Rule.

Step2: Apply L'Hopital's Rule first - time

Differentiate numerator and denominator. \(y = x+\pi\sec x\), \(y^\prime=1+\pi\sec x\tan x\); \(z = x^{2}-\pi^{2}\), \(z^\prime = 2x\). Now, \(\lim_{x
ightarrow\pi}\frac{1 + \pi\sec x\tan x}{2x}\). When \(x=\pi\), numerator \(1+\pi\sec\pi\tan\pi=1\), denominator \(2\pi\). So, the limit is \(\frac{1}{2\pi}\).

Answer:

C. \(\frac{1}{2\pi}\)