# What is the value of `lim_(x->oo) log(x + 1) - log x` . Do I need to use derivatives here ?

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### 3 Answers

It seems that the previous answer made a cut in some of the equations. But the term that is left is log (1+1/x) for which when x goes to infinity yields log 1 which is zero. So the whole limit asked for is zero.

In order to solve this problem, we need only to use properties of logarithms and properties of limits:

`lim_(x->oo)` log (x+1) - `lim_(x->oo) log (x)`

From the property of logs, the first term above can be written as

`lim_(x->oo) logx*(1+1/x)=lim_(x->oo){log x + log (1+1/x)}=lim_(x-> oo) log x + lim_(x-> oo)log(1+1/x)`

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Then since the term in `lim_(x->oo)log x`

cancels out we are left with

. For x going to infinity 1/x goes to zero and we are left with log 1 = 0.

The answer is therefore zero.

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The value of `lim_(x->oo) log (x+1) - log x` has to be determined.

`lim_(x->oo) log (x+1) - log x`

Use the property of logarithm `log a - log b = log(a/b)`

= `lim_(x->oo) log ((x+1)/x)`

= `lim_(x->oo) log (1 + 1/x)`

As `x->oo` , `1/x -> 0` , the given limit is equal to log 1 = 0 at x = 0.

**The value of the given limit is 0.**

There is no need to determine the derivative to find the limit in this case. Just the use of the property of logarithm and the fact that as the denominator in a fraction is increased, the value of the fraction decreases is sufficient.