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प्रश्न
Is the function defined by `f(x) = x^2 - sin x + 5` continuous at x = π?
उत्तर
f(x) = x2 - sin (x) + 5
If f(x) is continuous at x = `pi`, it implies:
`f (pi) = lim_(x -> pi^+) f(x) = lim _(x -> pi^-) f(x)`
`=> (pi^2 - sin(pi) + 5) = (pi^2 - sin (pi) + 5) = (pi^2 - sin (pi) + 5)`
`=> pi^2 + 5 = pi^2 + 5 = pi^2 + 5`
Which is true, i.e., f(x) is continuous at x = `pi`.
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