limit x-5 ((x^3-6*x^2+25)/(x-5) the answer of this question is 15 which we can determine by putting such values of of x which are close to 5 (but not exactly 5), for example first I started with these values 4.8,4.9,4.99 & -4.8,-4.9,-4.99. It's kind of a lengthy process, is there any quick process for such problem?

Question

limit x->5 ((x^3-6*x^2+25)/(x-5) the answer of this question is 15 which we can determine by putting such values of of x which are close to 5 (but not exactly 5), for example first I started with these values 4.8,4.9,4.99 & -4.8,-4.9,-4.99. It's kind of a lengthy process, is there any quick process for such problem?

anonymous · Answer

limit's are not just about substituting... we have to do some algebra first (if we can !)
you have denominator x-5  so substituting would be like  5-5=0   and we don't want that to happen.... so note that we can factor numerator ... and then we simplify...  here's how it goes !

$$\LARGE \lim_{x\to 5}\frac{x^3-6x^2+25}{x-5}=\lim_{x\to5}\frac{(x-5)(x^2-x-5)}{x-5}$$

$$\LARGE \lim_{x\to5}(x^2-x-5)=5^2-5-5=25-10=15$$
Hope this helps  :)

anonymous · Answer

what if we can't simplify the equation or if the equation is already simplified

anonymous · Answer

post an example.. let's see :(

anonymous · Answer

it's OK no problem thanks anyway

anonymous · Answer

ok... as you wish ;) Glad to help, you're welcome.