Find the limit x->infinity of: [ln(2 + x) - ln(1 + x)]

[MarkSA]

Hello,

1) Find the limit x->infinity of: [ln(2 + x) - ln(1 + x)] ... using l'hospital

It's not in infinity/infinity or 0/0 first.. it's in infinity - infinity which wont work for the rule. All I can think of doing is:
x-> infinity of: ln([2 + x]/[1 + x]) .. but is that considered to be in the form of f(x)/g(x) for l'hospital? And if so.. how would one take the derivative of the top and bottom?

Thanks

 

[pka]

\(\displaystyle \ln \left( {2 + x} \right) - \ln \left( {1 + x} \right) = \ln \left( {\frac{{2 + x}}{{1 + x}}} \right)\)

If F is continuous then
\(\displaystyle \lim _{x \to \infty } \ln (F(x)) = \ln \left( {\lim _{x \to \infty } F(x)} \right)\)

 

[Deleted member 4993]

Hello,

1) Find the limit x->infinity of: [ln(2 + x) - ln(1 + x)] ... using l'hospital

It's not in infinity/infinity or 0/0 first.. it's in infinity - infinity which wont work for the rule. All I can think of doing is:
x-> infinity of: ln([2 + x]/[1 + x]) .. but is that considered to be in the form of f(x)/g(x) for l'hospital? And if so.. how would one take the derivative of the top and bottom?

Thanks

[ln(2 + x) - ln(1 + x)]

= ln[x(2/x + 1)] - ln[(1/x + 1)x]

= ln[x] + ln(2/x + 1) - ln[x] - ln(1/x + 1)

= ln(2/x + 1) - ln(1/x + 1)

\(\displaystyle = ln\frac{1 + \frac{2}{x}}{1 + \frac{1}{x}}\)

Now continue....

 

[MarkSA]

Thanks I had forgotten that the ln can be pulled outside the limit... and the 1/x rule in limits too. answer becomes 0