The value of the following limit `lim_(ntooo) n(n+1)(ln(1+(1)/(n))-sin((1)/(n+1)))` is-

Options

A. `(1)/(2)`
B. `-(1)/(2)`
C. 0
D. does not exist

Answer

To solve the limit

\[ \lim_{n \to \infty} n(n+1)\left(\ln\left(1+\frac{1}{n}\right) - \sin\left(\frac{1}{n+1}\right)\right), \]

we will use Taylor series expansions for \(\ln(1+x)\) and \(\sin(x)\).

### Step 1: Expand \(\ln\left(1+\frac{1}{n}\right)\)

Using the Taylor series expansion around \(x = 0\):

\[ \ln(1+x) \approx x - \frac{x^2}{2} + O(x^3). \]

For \(x = \frac{1}{n}\):

\[ \ln\left(1+\frac{1}{n}\right) \approx \frac{1}{n} - \frac{1}{2n^2} + O\left(\frac{1}{n^3}\right). \]

### Step 2: Expand \(\sin\left(\frac{1}{n+1}\right)\)

Using the Taylor series expansion around \(x = 0\):

\[ \sin(x) \approx x - \frac{x^3}{6} + O(x^5). \]

For \(x = \frac{1}{n+1}\):

\[ \sin\left(\frac{1}{n+1}\right) \approx \frac{1}{n+1} - \frac{1}{6(n+1)^3} + O\left(\frac{1}{(n+1)^5}\right). \]

### Step 3: Substitute the expansions into the limit

Now, substituting the expansions into the limit expression:

\[ \ln\left(1+\frac{1}{n}\right) - \sin\left(\frac{1}{n+1}\right) \approx \left(\frac{1}{n} - \frac{1}{2n^2}\right) - \left(\frac{1}{n+1} - \frac{1}{6(n+1)^3}\right). \]

### Step 4: Simplify the expression

Now we simplify:

\[ \frac{1}{n} - \frac{1}{n+1} = \frac{(n+1) - n}{n(n+1)} = \frac{1}{n(n+1)}. \]

Thus,

\[ \ln\left(1+\frac{1}{n}\right) - \sin\left(\frac{1}{n+1}\right) \approx \frac{1}{n(n+1)} - \frac{1}{2n^2} + O\left(\frac{1}{n^3}\right). \]

### Step 5: Combine and multiply by \(n(n+1)\)

Now, we multiply by \(n(n+1)\):

\[ n(n+1)\left(\frac{1}{n(n+1)} - \frac{1}{2n^2} + O\left(\frac{1}{n^3}\right)\right) = 1 - \frac{n(n+1)}{2n^2} + O\left(\frac{1}{n}\right). \]

### Step 6: Simplify further

The term \(-\frac{n(n+1)}{2n^2}\) simplifies to:

\[ -\frac{(n+1)}{2n} = -\frac{1}{2} - \frac{1}{2n}. \]

As \(n \to \infty\), the \(O\left(\frac{1}{n}\right)\) term vanishes, and we are left with:

\[ 1 - \frac{1}{2} = \frac{1}{2}. \]

### Conclusion

Thus, the limit evaluates to:

\[ \lim_{n \to \infty} n(n+1)\left(\ln\left(1+\frac{1}{n}\right) - \sin\left(\frac{1}{n+1}\right)\right) = \frac{1}{2}. \]

### Final Answer

The value of the limit is \(\frac{1}{2}\).

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