Lim. f(x)=(x^2-x)/(2x^2 +5x-7) x → 1 Someone can help me ?

2 Answers
Mar 31, 2018

Answer:

See below.

Explanation:

#(x^2-x)/(2x^2 +5x-7) = (x(x-1))/(2(x+7/2)(x-1)) = x/(2(x+7/2))#

then

#lim_(x->1)(x^2-x)/(2x^2 +5x-7) = lim_(x->1)x/(2(x+7/2)) = 1/9#

Mar 31, 2018

Answer:

#lim_(xrarr1) (x^2-x)/(2x^2+5x-7)=color(red)(1/9)#

Explanation:

Note that in a case like the the object is always to reduce the numerator an denominator so that substitution of the limit value (in this case #1#) does not result in the undefined form: #0/0#.

After factoring
#color(white)("XXX")x^2-x=x(x-1)#
and
#color(white)("XXX")2x^2+5x-7=(2x+7)(x-1)#

Provided #x# is not quite equal to #1# (so that #x-1!=0#)
#lim_(xrarr1) (x^2-x)/(2x^2+5x-7)#

#color(white)("XXX")=lim_(xrarr1) (xcancel(""(x-1)))/((2x+7)cancel(""(x-1)))#

#color(white)("XXX")=1/(2 * 1 +7)#

#color(white)("XXX")=1/9#