- #1
Saracen Rue
- 150
- 10
Say we have two functions; ##f\left(x\right)=\frac{1}{e^x-1}## and ##g\left(x\right)=\ln \left(\frac{1}{x}+1\right)##. Let us find the limit of both functions as x approaches infinity;
##\lim_{x \rightarrow \infty} {f(x)} = \frac{1}{e^\infty-1} = \frac{1}{\infty} = 0## Therefore as ##x \rightarrow \infty##, ##f(x) \rightarrow 0##.
##\lim_{x \rightarrow \infty} {g(x)} = \ln \left(\frac{1}{\infty}+1\right) = \ln \left(0+1\right) = 0## Therefore as ##x \rightarrow \infty##, ##g(x) \rightarrow 0##.
Now, when we integrate ##f(x)## over the domain ##[1, \infty)##, it works perfectly fine and has a result of
0.45868. This makes sense because the area enclosed under the graph is constantly getting smaller due to ##f(x) \rightarrow 0##.
However, when I try to integrate ##g(x)## over the same domain my calculator won't give me an answer. ##g(x)## approaches 0 just as ##f(x)## does, meaning that the enclosed area should increasingly get smaller also until it reaches the point where the area incriminates being added become negligible. So why can't I get a numerical answer while integrating ##g(x)##?
##\lim_{x \rightarrow \infty} {f(x)} = \frac{1}{e^\infty-1} = \frac{1}{\infty} = 0## Therefore as ##x \rightarrow \infty##, ##f(x) \rightarrow 0##.
##\lim_{x \rightarrow \infty} {g(x)} = \ln \left(\frac{1}{\infty}+1\right) = \ln \left(0+1\right) = 0## Therefore as ##x \rightarrow \infty##, ##g(x) \rightarrow 0##.
Now, when we integrate ##f(x)## over the domain ##[1, \infty)##, it works perfectly fine and has a result of
0.45868. This makes sense because the area enclosed under the graph is constantly getting smaller due to ##f(x) \rightarrow 0##.
However, when I try to integrate ##g(x)## over the same domain my calculator won't give me an answer. ##g(x)## approaches 0 just as ##f(x)## does, meaning that the enclosed area should increasingly get smaller also until it reaches the point where the area incriminates being added become negligible. So why can't I get a numerical answer while integrating ##g(x)##?