How do you find a value k such that the limit exists given #(x^2+8x+k)/(x+2)#?

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Answer 1 · 2017-08-18T00:48:48

If we define:
# L(alpha) = lim_(x rarr alpha) (x^2+8x+k)/(x+2) #

If #alpha=-2# the limit exists provided #k=12 => L(-2) = 4#

For any other value of alpha, the limit always exist, #k# is arbitrary, and

# L(alpha) = (alpha^2+8alpha+k)/(alpha+2) #

The question is incomplete as the value of #x# is not specified for the limit.

Let us consider the function:

# L(alpha) = lim_(x rarr alpha) (x^2+8x+k)/(x+2) #

The limit function is not defined when the denominator is zero, ie when #x+2=0=> x=-2#

So let us consider the case #alpha=-2#, in which case we examine:

# L(-2) = lim_(x rarr -2) (x^2+8x+k)/(x+2) #

This limit will exist iff the numerator also has a factor of #(x+2)#, along with another factor (x+beta), say, This would require:

# (x+beta)(x+2) -= x^2+8x+k #
# =>x^2+(2+beta)x+2beta = x^2+8x+k #

If we equate coefficients then we get:

# x^1 : 2+beta = 8 => beta = 6 #
# x^0 : 2beta = k => k=12 #

Hence, we have a factor that will cancel with #beta=6,k=12# givings

# L(-2) = lim_(x rarr -2) (x^2+8x+12)/(x+2) #
# " " = lim_(x rarr -2) ((x+6)(x+2))/(x+2) #
# " " = lim_(x rarr -2) (x+6) #
# " " = 4 #

For any other value of #alpha#, the limit will always exist, ie #k# is arbitrary, and

# L(alpha) = lim_(x rarr alpha) (x^2+8x+k)/(x+2) #
# " " = (alpha^2+8alpha+k)/(alpha+2) #