How do you find the limit of #(sqrtx)/ln(x+1)# as x approaches 0+?

1 Answer
Nathan L.
Aug 4, 2017

#lim_(xrarr0^+) [(sqrtx)/(ln(x+1))] = color(blue)(oo#

Explanation:

Without using special rules...

To find the limit of the function as #x# approaches #0# from the right, we can plug in successively smaller positive numbers that get closer and closer to zero (i.e. #0.1#, #0.01#, #0.001#, etc.)

#0.1:" "# #(sqrt(0.1))/(ln(0.1+1)) ~~ ul(3.318#

#0.01:" "# #(sqrt(0.01))/(ln(0.01+1)) ~~ ul(10.050#

#0.001:" "# #(sqrt(0.001))/(ln(0.001+1)) ~~ ul(31.639#

#...#

#10^-15:" "# #(sqrt(10^-15))/(ln(10^-15+1)) ~~ ul(3.162xx10^7#

We can infer that the numbers get larger and larger as we go onward (which they do).

Also notice that in this limit, the denominator approaches #ln(1)#, which equals #0#, and as you divide a number by smaller and smaller numbers, the result approaches infinity, so we can conclude that

#lim_(xrarr0^+) [(sqrtx)/(ln(x+1))] = color(blue)(oo#

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