# If limx→0 g(x)=0, and |h(x)| ≤M, for all x, prove that limx→0[g(x) h(x)]=0.

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### 1 Answer

Given `lim_(x->0)g(x)=0` and `|h(x)|<=M` for all x, show `lim_(x->0)[g(x)h(x)]=0`

We need to show that for any `epsilon>0` there exists a `delta>0` such that `0<|x-0|<delta==>|g(x)h(x)-0|<epsilon` . This is equivalent to showing `|g(x)M|<epsilon` since `|h(x)|<=M`

` `(1) `lim_(x->0)g(x)=0==>`

there exists a `delta>0` such that `0<|x-0|<delta` implies that `|g(x)-0|<epsilon` for all `epsilon>0` .

(2) ``Given `epsilon>0` ; let `delta=epsilon/M` .

Then `0<|x-0|<epsilon/M` implies that `|g(x)-0|<epsilon/M` since the limit of g(x) exists

Then `M|g(x)|<epsilon==>|g(x)M|<epsilon`

and `|g(x)h(x)-0|<epsilon` .

This is true for all `epsilon` , so the limit is 0.

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