What is the limit of # (x^2 − 5x)/(x^2 − 4x − 5)# as x approaches 5?

Question

Elijah Abram · Accepted Answer

#5/6# #x^2-5x=x(x-5)# and #x^2-4x-5=(x-5)(x+1)# thus we get #(x*(x-5))/((x-5)(x+1))=x/(x+1)# for #xne 5#

Luke Alvarez · Answer

#lim_(xrarr5)(x^2 − 5x)/(x^2 − 4x − 5)=lim_(xrarr5)[2x-5]/[2x-4]=[10-5]/[10-4]=5/6# #lim_(xrarr5)(x^2 − 5x)/(x^2 − 4x − 5)=[25-25]/[25-25]=0/0# #"L'hospital Rule"# since the direct compensation product equal #0/0# we will use L'hospital Rule #color(red)[lim_(trarra)(f'(x))/(g('x))]# #f(x)=x^2 − 5x# #f'(x)2x-5# #g(x)=x^2 − 4x − 5# #g'(x)=2x-4# #lim_(xrarr5)(x^2 − 5x)/(x^2 − 4x − 5)=lim_(xrarr5)[2x-5]/[2x-4]=[10-5]/[10-4]=5/6#

Isaiah Allen · Answer

#\frac{5}{6}#

You can factor #x# in the numerator:

#x^2-5x = x(x-5)#

You can factor the denominator by finding the roots of the polynomial. In this case, we can use the sum/product method: if the coefficient of #x^2# is #1#, then we can write the equation as #x^2-sx+p#, where #s# is the sum of the solutions and #p# is their product.

So, we're looking for two numbers #x_1# and #x_2# such that:

#x_1+x_2 = 4# #x_1x_2 = -5#

So, #x_1 = -1# and #x_2 = 5#

And finally #x^2-4x-5=(x+1)(x-5)#

The fraction becomes

#\frac{x^2-5x}{x^2-4x-5}=\frac{x(x-5)}{(x+1)(x-5)}=\frac{x}{x+1}#

So, when #x# approaches #5#, the limit is #\frac{5}{6}#

Cameron Alexander · Answer

undefined The limit of (x^2 − 5x)/(x^2 − 4x − 5) as x approaches 5 is 5/6.