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Evaluate: `=lim_(zrarr0) m(5+3x+...+x)/(x^(2))` |
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Answer» `underset(xrarroo)"lim"(1+2+3+.....+x)/(x^(2)) ((oo)/(oo))` `=underset(xrarroo)"lim"((1)/(2)x(X+1))/(x^(2))` `=underset(xrarroo)"lim"(1)/(2)(1+(1)/(x))` `=underset(zrarr0)"lim"(1)/(2)(1+z) ("put" (1)/(x)=z)` `=(1)/(2)(1+0)=(1)/(2)`. |
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