How do you determine the limit of #(x+5)/(x-2)# as x approaches 2+?

Question

Isaiah Allen · Accepted Answer

The limit tends to #oo#.

We have to solve #lim_(xrarr2^+)(x+5)/(x-2)# Directly inputting will not work, as the denominator will give #0#, and division by #0# is not possible. So we make #x# get closer and closer to #2#. For example, we take #2.1#, then #2.01#, #2.001#, and so on. For #2.1#: #(2.1+5)/(2.1-2)=7.1/0.1=71# For #2.01#: #(2.01+5)/(2.01-2)=7.01/0.01=701# And so on. We see a pattern. The closer we get to #x=2#, the value of the limit increases dramatically. This is because the division of a number by an extremely small one gives us an extremely large answer. Therefore, we can say: #lim_(xrarr2)(x+5)/(x-2)=+-oo# However, the questions asks us to get to two from values larger than it, as shown in the examples. This gives us a positive denominator and numerator, showing us that, when #xrarr2^+#, #-oo# is not valid. So we say #lim_(xrarr2^+)(x+5)/(x-2)=oo#

Scarlett Allison · Answer

undefined To determine the limit of (x+5)/(x-2) as x approaches 2+, we substitute the value 2 into the expression. However, since the denominator becomes zero at x=2, we cannot directly substitute the value. In this case, we can use the concept of one-sided limits.

To find the limit as x approaches 2+ (from the right side), we evaluate the expression for values of x that are slightly greater than 2. By substituting x=2.1, x=2.01, x=2.001, and so on, we observe that the expression approaches positive infinity. Therefore, the limit of (x+5)/(x-2) as x approaches 2+ is positive infinity.