How do you evaluate #( 5x^9-4x^2-2x+1)/(x^7-x^6+x-1)# as x approaches 1?

#lim_(x->1) (5x^9-4x^2-2x+1)/(x^7-x^6+x-1)#

#=lim_(x->1) (color(red)(cancel(color(black)((x-1))))(5x^8+5x^7+5x^6+5x^5+5x^4+5x^3+5x^2+x-1))/(color(red)(cancel(color(black)((x-1))))(x^6+1))#

#=lim_(x->1) (5x^8+5x^7+5x^6+5x^5+5x^4+5x^3+5x^2+x-1)/(x^6+1)#

#=(5+5+5+5+5+5+5+1-1)/(1+1)#

#=35/2#

#color(white)()#
Alternative method

Alternatively you can use L'Hôpital's rule in the form:

If the following hold:

• #f(x)# and #g(x)# are differentiable on an open interval containing #c#, except possibly at #c#.

• #lim_(x->c) f(x) = lim_(x->c) g(x) = 0#

• #lim_(x->c) (f'(x))/(g'(x))# exists

Then:

#lim_(x->c) f(x)/g(x) = lim_(x->c) (f'(x))/(g'(x))#

#color(white)()#
For our example, let:

#f(x) = 5x^9-4x^2-2x+1#

#g(x) = x^7-x^6+x-1#

#c = 1#

Then:

• #f(x)# and #g(x)# are differentiable on an open interval containing #c=1# with:

#f'(x) = 45x^8-8x-2#

#g'(x) = 7x^6-6x^5+1#

• #lim_(x->1) f(x) = lim_(x->1) g(x) = 0#

• #lim_(x->1) (f'(x))/(g'(x)) = (45-8-2)/(7-6+1) = 35/2# exists

So:

#lim_(x->1) f(x)/g(x) = lim_(x->1) (f'(x))/(g'(x)) = 35/2#