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\operatorname { lim } _ { h \rightarrow 0 } \frac { \operatorname { sin } ^ { 2 } ( x + h ) - \operatorname { sin } ^ { 2 } x } { h } |
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Answer» if we put h=0 we will get 0/0 formso using L'hospital rulelim h->0 (2sin(x+h)cos(x+h)-2sinxcosx)/1=2sinxcosx-2sinxcosx=0 |
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