高数求极限 要详解lim (ln tanx -ln x)/(x^2) (x->0)

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高数求极限要详解lim(lntanx-lnx)/(x^2)(x->0)高数求极限要详解lim(lntanx-lnx)/(x^2)(x->0)高数求极限要详解lim(lntanx-lnx)/(x^2)(

高数求极限 要详解lim (ln tanx -ln x)/(x^2) (x->0)
高数求极限 要详解
lim (ln tanx -ln x)/(x^2) (x->0)

高数求极限 要详解lim (ln tanx -ln x)/(x^2) (x->0)
lim【x→0】(lntanx-lnx)/(x^2)
=lim【x→0】ln[(tanx)/x]/(x^2)
=lim【x→0】ln[(sinx/cosx)/x]/(x^2)
=lim【x→0】ln(1/cosx)/(x^2)
=lim【x→0】(1/cosx-1)/(x^2)
=lim【x→0】(1-cosx)/(cosx·x^2)
=lim【x→0】[(x^2)/2]/(x^2)
=1/2
答案;:1/2

趋于无穷大
也不对
正确答案是
(ln tanx -ln x)/(x^2)=ln(tanx/x)/(x^2)
=ln(tanx/x-1+1)/ (x^2 )
=(tanx/x-1)/ (x^2 )
...

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趋于无穷大
也不对
正确答案是
(ln tanx -ln x)/(x^2)=ln(tanx/x)/(x^2)
=ln(tanx/x-1+1)/ (x^2 )
=(tanx/x-1)/ (x^2 )
=(tanx-x)/ (x^3 )
=1/3
用等价无穷小替换,其中(tanx-x)/ (x^3 )可用洛比达法则算出。ln(1+x)~x

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