Please answer it!!!
please dont write the x derivative way! go the normal way...
This gives an indeterminate form (0/0), so use L'Hôpital's rule:
lim[x→π] (π^2 - x^2)/sin(x)
= lim[x→π] [d(π^2 - x^2)/dx]/{d[sin(x)]/dx}
= lim[x→π] -2x/cos(x)
= -2π/(-1)
= 2π
First note that
sin(x - π) = -sin(x).
Therefore,
(π^2 - x^2) / sin(x)
= ( -1 (x + π) (x - π) ) / ( -sin(x - π) )
= (x + π) * (x-π)/sin(x-π)
Now, as x -> π, we have
x-π -> 0
and therefore
(x-π) / sin(x-π) -> 1.
Now, taking the limit of
(x + π) * (x-π)/sin(x-π),
we have
(π + π) * 1 = 2π.
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Verified answer
This gives an indeterminate form (0/0), so use L'Hôpital's rule:
lim[x→π] (π^2 - x^2)/sin(x)
= lim[x→π] [d(π^2 - x^2)/dx]/{d[sin(x)]/dx}
= lim[x→π] -2x/cos(x)
= -2π/(-1)
= 2π
First note that
sin(x - π) = -sin(x).
Therefore,
(π^2 - x^2) / sin(x)
= ( -1 (x + π) (x - π) ) / ( -sin(x - π) )
= (x + π) * (x-π)/sin(x-π)
Now, as x -> π, we have
x-π -> 0
and therefore
(x-π) / sin(x-π) -> 1.
Now, taking the limit of
(x + π) * (x-π)/sin(x-π),
we have
(π + π) * 1 = 2π.