Question #c0c50

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Answer 1 · 2017-04-15T07:20:43

$6.$

Let us solve the Problem without using L'Hospital's Rule.

First, we simplify the given rational function :

${(1+x)(1+2x)(1+3x)-1}/x$

$=[{1^3+(x+2x+3x)*1^2+(x*2x+2x*3x+3x*x)*1^1+x*2x*3x}-1]/x$

$=(1+6x+11x^2+6x^3-1)/x$

$={x(6+11x+6x^2)}/x.$

$"Therefore, the Reqd. Limit="lim_(x to 0) (6+11x+6x^2)=6.$

Enjoy Maths.!