求lim(x->2)[(x+2)/(x^2+3x-10)]的极限.答案是无穷大(横着的8的符号).我想要具体的步骤!希望懂的人帮帮忙.还有一题:lim(x->0)[1-(1+2^2)]/x^2 (1+2^2)这个是有根号的,我手机打不出来,将就看了,知道就行.这

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求lim(x->2)[(x+2)/(x^2+3x-10)]的极限.答案是无穷大(横着的8的符号).我想要具体的步骤!希望懂的人帮帮忙.还有一题:lim(x->0)[1-(1+2^2)]/x^2 (1+2^2)这个是有根号的,我手机打不出来,将就看了,知道就行.这
求lim(x->2)[(x+2)/(x^2+3x-10)]的极限.
答案是无穷大(横着的8的符号).我想要具体的步骤!希望懂的人帮帮忙.还有一题:lim(x->0)[1-(1+2^2)]/x^2 (1+2^2)这个是有根号的,我手机打不出来,将就看了,知道就行.这题的答案是 (-1/2)

求lim(x->2)[(x+2)/(x^2+3x-10)]的极限.答案是无穷大(横着的8的符号).我想要具体的步骤!希望懂的人帮帮忙.还有一题:lim(x->0)[1-(1+2^2)]/x^2 (1+2^2)这个是有根号的,我手机打不出来,将就看了,知道就行.这
说明：第二题有错误,正确的是lim(x->0)[(1-√(1+x²))/x²].
lim(x->2)[(x+2)/(x²+3x-10)]=lim(x->2)[(x+2)/((x-2)(x+5))]
=(2+2)/((2-2)(2+5))
=4/(0*7)
=4/0
=∞；
lim(x->0)[(1-√(1+x²))/x²]=lim(x->0)[((1-√(1+x²))(1+√(1+x²)))/(x²(1+√(1+x²)))] (分子有理化)
=lim(x->0)[(1-(1+x²))/(x²(1+√(1+x²)))]
=lim(x->0)[(-x²)/(x²(1+√(1+x²)))]
=lim(x->0)[(-1)/(1+√(1+x²))]
=[(-1)/(1+√(1+0²))]
=-1/2.

4/0->∞
第二个没法将就着看

lim(x->2)[(x+2)/(x^2+3x-10)] = 4/0 = 无穷大

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