How do you determine the limit of #1/(x²+5x-6)# as x approaches -6?

Question

2027 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-21T21:06:22

DNE-does not exist

#lim_(x->-6) 1/((x+6)(x-1))#

#=1/(0*-7)#

#=1/0#

#DNE#

Answer 2 · 2016-04-21T22:20:03

The limit does not exist. Look at the signs of the factors.

Let #f(x) = 1/(x^2+5x-6) = 1/((x+6)(x-1))#

Not that as #xrarr-6#, we have #(x-1) rarr -7#

From the left

As #xrarr-6^-#, the factor #(x+6)rarr0^-#, so #f(x)# is positive and increasing without bound.

#lim_(xrarr-6^-)f(x) = oo#

From the right

As #xrarr-6^+#, the factor #(x+6)rarr0^+#, so #f(x)# is negative and increasing without bound.

#lim_(xrarr-6^+)f(x) = -oo#

Two sided

#lim_(xrarr-6)f(x)# does not exist.