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Solution to the true questions of mathematical analysis postgraduate entrance examination
Because f is continuous on the closed interval, there is m >；; 0, so | f0 (x) | < M holds for all X ..

therefore

| f 1(x)| & lt； = M|x|，

...

| f(n)(x)| & lt； = M|x|^n/n！

....

Because m | x | n/n! & lt= Ma^n/n！ -& gt； 0

So the conclusion is established.

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