lim x-->0+ f(x), where f(x)= x ln x
values for x are 1,0.5,0.1,0.05,0.01,0.005,0.001
when i plugged in 1 i got 0 then -.34657 then -.23025 ................... leading up to -0.00690
what is the value of the limit is it approaching 0?? and if it is approaching 0 then why is the value for 1 --- 0 ?? :S
Using L'Hopital's rule, with
f(x) = x lnx= lnx /(1/x),
Limit as x->0 = Lim[(1/x)/(-1/x^2)]
= Lim(-x) = 0
There is no reason why the value of f(1) cannot also be the limit of f(x) as x-> 0
f(x) has a minimum between 0 and 1, at x = 1/e. The minimum is -0.3678
To find the limit as x approaches 0+ of the function f(x) = x ln x, we can plug in smaller and smaller values of x that are approaching 0 from the positive side to see how the function behaves.
When you plugged in 1, you obtained a value of 0 for f(x) because ln(1) = 0. In this case, f(1) = 1 * ln(1) = 1 * 0 = 0.
However, as you continued to plug in smaller values for x, you noticed that the value for f(x) became negative and decreased. This indicates that the function is approaching a negative value as x approaches 0 from the positive side.
To determine the actual value of the limit, we can observe the behavior of the function as x gets closer to 0. Let's analyze the values you provided:
- f(1) = 0
- f(0.5) = 0.5 * ln(0.5) ≈ -0.34657
- f(0.1) = 0.1 * ln(0.1) ≈ -0.23025
- f(0.05) = 0.05 * ln(0.05) ≈ -0.17639
- f(0.01) = 0.01 * ln(0.01) ≈ -0.04605
- f(0.005) = 0.005 * ln(0.005) ≈ -0.01828
- f(0.001) = 0.001 * ln(0.001) ≈ -0.00690
As you can see, the values are decreasing and approaching a negative value as x gets closer to 0. This suggests that the limit as x approaches 0+ of f(x) is approximately -0.00690.