How do you find the limit of # (x^2 + 3x - 10)# as x approaches #2^+#?

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Answer 1 · 2016-05-16T15:39:18

#lim_(xrarr2^+)(x^2+3x-10) = 0#.

As #xrarr2# (from either side), #x^2 rarr4# and #3xrarr6#, so #lim_(xrarr2)(x^2+3x-10) = 4+6-10 = 0#.

More information

It is possible that this is not all the information needed.

The limit is #0#, but in some situations we also need to know whether the values are approaching #0# through positive or negative numbers.

#x^2+3x-10 = (x-2)(x+5)#

As #xrarr2^+#, both of these factors are positive, so we might write

#lim_(xrarr2)(x^2+3x-10) = 0^+#.

We would need this if, for example, we wanted to evaluate

#lim_(xrarr2^+)(1-x)/(x^2+3x-10)#.

The numerator approaches #-1# and the denominator approaches #0# but is positive, so the ratio is decreasing without bound. We write

#lim_(xrarr2^+)(1-x)/(x^2+3x-10) = -oo#.