How do you find the limit of #(sqrtx)/ln(x+1)# as x approaches 0+?
1 Answer
Explanation:
Without using special rules...
To find the limit of the function as
#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
#lim_(xrarr0^+) [(sqrtx)/(ln(x+1))] = color(blue)(oo#