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5 votes
5 votes
Why f(
\pi) = 0 and f '(
\pi) = 3

When
\lim_(x \to \\\pi ) (f(x))/(x-\pi) = 3

User Gullbyrd
by
2.6k points

1 Answer

27 votes
27 votes

Answer:

application of L'Hopital's rule to the presumed indeterminate form yields this conclusion

Explanation:

This is a reverse application of L'Hopital's rule for determining the limits involving indeterminate forms.

When the expression evaluated at the limit is 0/0, then L'Hopital's rule tells you the limit can be found from n'/d', where n and d are the numerator and denominator of the original expression, respectively.

We can see that x-π = 0 at x=π, so we assume that f(π) = 0 as well, and the expression n/d = f(x)/(x-π) evaluates to the indeterminate form 0/0.

The derivatives are ...

n' = f'(x)

d' = 1

Then we have the limit as ...

lim{x→π) = n'/d' = f'(π)/1 = 3 ⇒ f'(π) = 3

The conclusion f(π)=0 and f'(π)=3 follows from L'Hopital's Rule.

User Tarek Badr
by
2.7k points