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

Question

Christopher Agresta · Accepted Answer

#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#.

Paisley Johnson · Answer

undefined To find the limit of a function as x approaches a specific value, we can substitute that value into the function. In this case, we substitute x = 2 into the function (x^2 + 3x - 10). By doing so, we get (2^2 + 3(2) - 10), which simplifies to (4 + 6 - 10) = 0. Therefore, the limit of (x^2 + 3x - 10) as x approaches 2^+ is 0.