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Mathematical limit questions for postgraduate entrance examination
1, the original formula = limx → -∞100x/[(√ (x2+100)-x].

= limx→-∞ 100/[-√( 1+ 100/x^2)- 1]

= 100/[-√( 1+0)- 1]

=-50;

2. the first picture: limx →∞ e * (1+1/x) (-x)-1= e * (1/e)-1= 0,

That is, e * (1+1/x) (-x)-1is infinitesimal,

-e ln [e * (1+1/x) (-x)]-1is infinitesimal,

e^x- 1~x,

——》e^ln[e*( 1+ 1/x)^(-x)]- 1~ln[e*( 1+ 1/x)^(-x)];

The second picture: I don't understand it either.