1.

Evaluate: `=lim_(zrarr0) m(5+3x+...+x)/(x^(2))`

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)`.


Discussion

No Comment Found