EXAMPLE OF USE OF THEOREMS ON LIMITS

FIND: limx --> 0 [x2sin(1/x)]/sinx

We know that

limx --ɬ (sinx)/x = 1

? limx --> 0 [x2sin(1/x)]/sinx =

limx --> 0 [xsin(1/x)]/(sinx/x) = limx --> 0 [xsin(1/x)]/1

Now -1<sin(1/x)ə and limx --> 0 x = 0, so

limx --> 0 [xsin(1/x)]/1= 0 * limx --> 0 sin(1/x)= 0

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