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How do you determine the limit of #(x+2)/(x+3)# as x approaches #-3^+#?

Answer 1

# -oo#

#lim_{x \to -3^+}(x+2)/(x+3)#

if we just plug in #x = -3#, we can see that it is #2/oo#. the graph shows that #lim_{x \to -3^+}(x+2)/(x+3) = - oo#

to see this, let #x = -3 + \epsilon# {ie just to right of x = -3], with #0 < \epsilon < < 1# we have

#(-3 + \epsilon + 2)/(-3 + \epsilon + 3)#
#= ( -1 + \epsilon )/(\epsilon)#
#= ( -1)/(\epsilon) + (\epsilon )/(\epsilon)#
#= -( 1)/(\epsilon) + 1#

we see that the dominant term for small #\epsilon# is #= -( 1)/(\epsilon) #, which is negative; and #lim_{\epsilon \to 0} -( 1)/(\epsilon) + 1 = -oo#

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Answer 2

Emilia Aguilar

To determine the limit of (x+2)/(x+3) as x approaches -3^+, we substitute -3 into the expression. This gives us (-3+2)/(-3+3), which simplifies to -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.