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

Answer 1

Christopher Agresta

DNE-does not exist

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

#=1/(0*-7)#

#=1/0#

#DNE#

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Answer 2

Charles Anctil

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.

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Answer 3

Ezra Adams

To determine the limit of 1/(x²+5x-6) as x approaches -6, we can substitute -6 into the expression and simplify. By substituting -6 for x, we get 1/((-6)²+5(-6)-6). Simplifying further, we have 1/(36-30-6), which becomes 1/(0). Since division by zero is undefined, the limit does not exist.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.