How Do You Solve the Limit of sin(5x)/(x-π) as x Approaches π?

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SUMMARY

The limit of sin(5x)/(x-π) as x approaches π evaluates to -5. The discussion highlights the application of the limit property lim(x→0) sin(x)/x = 1 by substituting t = x - π, transforming the limit to lim(t→0) -sin(5t)/t. This approach effectively resolves the indeterminate form 0/0 encountered initially. The final result confirms the limit as -5, demonstrating the utility of trigonometric limit properties in calculus.

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Homework Statement



evaluate the limit: Limit x--> \Pi
sin(5x) / x-\Pi

Homework Equations





The Attempt at a Solution



I get 0/0 since I figure I can't cancel anything out...I think there's another way of solving it I just don't know what way it is...

Thanks for the help.
 
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Are you familiar with the following limit ?

\lim_{x \to 0} \frac{sin \left( x \right)}{x} = 1
 
Last edited:


Yes so if I multiplied top and bottom by 5. I would get 5sin(x) / 5x - 5 \pi wouldn't that give me something like 5/-5\pi ?
 


hpthinker said:
Yes so if I multiplied top and bottom by 5. I would get 5sin(x) / 5x - 5 \pi wouldn't that give me something like 5/-5\pi ?

I misread your question.

Sorry about that.Why don't you do the following t = x - \pi

Your limit becomes \lim_{t \to 0} \frac{ sin(5(t- \pi))}{t} = \lim_{t \to 0} \frac{-sin(5t)}{t}
 
Last edited:

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