求lim(x→π/4) (sin2x-cos2x-1)/(cosx-sinx)的极限

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求lim(x→π/4) (sin2x-cos2x-1)/(cosx-sinx)的极限

求lim(x→π/4) (sin2x-cos2x-1)/(cosx-sinx)的极限
求lim(x→π/4) (sin2x-cos2x-1)/(cosx-sinx)的极限

求lim(x→π/4) (sin2x-cos2x-1)/(cosx-sinx)的极限
[[1]]
∵cos2x=2cos²x-1.
∴cos2x+1=2cos²x.
且sin2x=2sinxcosx.
∴分子=2sinxcosx-2cos²x
=2cosx(sinx-cosx)
∴原式=-2cosx.
∴当x--->π/4时,
原式-->-2cos(π/4)=-√2