How do you find the limit of #sinx/(x-pi)# as #x->pi#?

2 Answers
Douglas K. · Steve M
Oct 26, 2016

Use L'Hôpital's rule #lim_(x→pi)sin(x)/(x - pi)= lim_(x→pi)cos(x) = -1#

Explanation:

Use L'Hôpital's rule

The derivative of the numerator is #cos(x)#

The derivative of the denominator is 1 so we will not write it.

#lim_(x→pi)sin(x)/(x - pi) = lim_(x→pi)cos(x) = -1#

NOTE
The question was posted in "Determining Limits Algebraically" , so the use of L'Hôpital's rule is NOT a suitable method to solve the problem

Jim H
Oct 27, 2016

I see that substitution gets us the indeterminate form #0/0#.

Explanation:

A limit like this reminds me that #lim_(thetararr0) sintheta/theta = 1#

And that makes me wonder if can use #sin(x-pi)# somehow.

The difference formula for the sine gets us:

#sin(x-pi) = sinx cos pi - cosx sinpi = sinx(-1) - cosx (0) = -sinx#

Therefore we can replace #sinx# by #-sin(x-pi)#

#lim_(xrarrpi)sinx/(x-pi) = lim_(xrarrpi) (-sin(x-pi))/(x-pi)#

# = - lim_(theta rarr0) sintheta/theta# #" "# (with #theta = x-pi#)

# = -1#

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