How do you evaluate the limit of #lim (-4x^2-3x+6)# as #x->2#?

Answer 1

#lim_(x rarr 2) (-4x^2-3x+6)# = #-16#

#lim_(x rarr 2) (-4x^2-3x+6)#

#= (-4*2^2 - 3*2 + 6)#

#=(-16 -6 +6)#

#=-16# The function is continuous at x = 2, where the function evaluates to #-16# and #lim_(x rarr 2^-)(-4x^2-3x+6)# = #lim_(x rarr 2 ^+)(-4x^2-3x+6)# = #-16#

Hence Proved.

graph{-4x^2 -3x + 6 [1.092, 2.778, -16.412, -15.569]}

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Answer 2

Christopher Agresta

To evaluate the limit of lim (-4x^2-3x+6) as x approaches 2, we substitute the value of 2 into the expression. Thus, the limit is equal to -4(2)^2 - 3(2) + 6, which simplifies to -16 - 6 + 6. Further simplification gives us -16. Therefore, the limit is -16.

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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.

How do you evaluate the limit of #lim (-4x^2-3x+6)# as #x-&gt;2#?

How do you evaluate the limit of #lim (-4x^2-3x+6)# as #x->2#?