Which is the limit of sin^2x/x?

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Answer 1 · 2018-04-06T23:22:13

=0

$lim_(x->0) (sin^2x)/x$

---- $lim_(x->0) (sinx)/x=1$

multiply by

$lim_(x->0) (sinx.sinx)/x=lim_(x->0) (x)/x(sinx.sinx)/x$

$lim_(x->0) (x)/x(sinx.sinx)/x=lim_(x->0)x((sinx.sinx))/(x.x)=lim_(x->0)(sinx/x)(sinx/x)(x)$

$lim_(x->0)(sinx/x)(sinx/x)(x)=1.1.x=x$

$lim_(x->0) (sin^2x)/x=lim_(x->0) x$

$lim_(x->0) x=0$