作业帮设f'(x0)=3,利用导数定义计算极限.1)lim h→0 [f(x0+2h)-f(x0)] / h ;lim h→0 [f(x0)-f(x0-h)]/h
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![设f'(x0)=3,利用导数定义计算极限.1)lim h→0 [f(x0+2h)-f(x0)] / h ;lim h→0 [f(x0)-f(x0-h)]/h](/uploads/image/z/3857537-65-7.jpg?t=%E8%AE%BEf%27%28x0%29%3D3%2C%E5%88%A9%E7%94%A8%E5%AF%BC%E6%95%B0%E5%AE%9A%E4%B9%89%E8%AE%A1%E7%AE%97%E6%9E%81%E9%99%90.1%29lim+h%E2%86%920+%5Bf%28x0%2B2h%29-f%28x0%29%5D+%2F+h+%3Blim+h%E2%86%920+%5Bf%28x0%29-f%28x0-h%29%5D%2Fh)
设f'(x0)=3,利用导数定义计算极限.1)lim h→0 [f(x0+2h)-f(x0)] / h ;lim h→0 [f(x0)-f(x0-h)]/h
设f'(x0)=3,利用导数定义计算极限.1)lim h→0 [f(x0+2h)-f(x0)] / h ;lim h→0 [f(x0)-f(x0-h)]/h
设f'(x0)=3,利用导数定义计算极限.1)lim h→0 [f(x0+2h)-f(x0)] / h ;lim h→0 [f(x0)-f(x0-h)]/h
（1）=lim h→0 [f(x0+2h)-f(x0)] / 2h *2=6
（2）=lim h→0 [f(x0)-f(x0-h)]/-h *-1=-3