Limit question

why is lim (x->0) of (pi*x) / (sin(pi*x)) = 1?????

The Maclaurin series for sin(z)=z-z^3/(3!)+...
let z = pi*x
the lim as x tends to 0 lets you ignore the series except the first term

Reply 2

InOrbit

Three main ways of doing this: Taylor series, L'Hopital's rule and geometric.

For the last way, can you try to prove by use of a diagram that

\sin z \leq z \leq \tan z

for

z \in (0,\pi/2)

. The areas of a triangle and sectors will be helpful here. Then manipulate it to get the limit of

\sin z / z

Reply 3

Raiden10

I always heard the sky is the limit.

Reply 4

Zacken

This is a bit fiddly. It depends on the way you define your trigonometric functions.

For example, both L'Hopitals Rule and the Taylor series suggested above requires that you know how to differentiate

\sin z

, but knowing how to differentiate

\sin z

means you need to know that

\lim_{z \to 0 } \frac{\sin z}{z}

anyway, so it's all circular (depending on your starting definitions).

Reply 5

B_9710

Original post by Alex:

Three main ways of doing this: Taylor series, L'Hopital's rule and geometric.

For the last way, can you try to prove by use of a diagram that

\sin z \leq z \leq \tan z

for

z \in (0,\pi/2)

. The areas of a triangle and sectors will be helpful here. Then manipulate it to get the limit of

\sin z / z

For the limit to exist it has to exist from the positive and negative direction so you have to consider both 1st and 4th quadrant.

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Limit question

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