How do you find the limit of #4 ((sinxcosx- sinx)/( x^2))# as x approaches 0?

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Answer 1 · 2015-10-19T11:45:30

Rewrite and use the fundamental trigonometric limits.

Recall that

#lim_(hrarr0)sinh/h = 1# and

#lim_(hrarr0) (cosh-1)/h = 0#.

We want

#lim_(xrarr0)4 ((sinxcosx- sinx)/( x^2))#

#4 ((sinxcosx- sinx)/( x^2)) = 4(sinx(cosx-1))/(x*x)#

# =4(sinx/x)((cosx-1)/x)#

So,

#lim_(xrarr0)4 ((sinxcosx- sinx)/( x^2)) = lim_(xrarr0)4(sinx/x)((cosx-1)/x) #

# = 4(1)(0) = 0#