To determine the value of the limit \(\mathop {\lim }\limits_{x \to \infty } \frac{{x {\:}ln\left( x \right)}}{{1 + {x^2}}}\), we first need to analyze the form of the expression as \(x\) approaches infinity.

Limit Indeterminate Form

As \(x \to \infty\):

• The numerator, \(f\left( x \right) = x \ln\left( x \right)\), approaches \(\infty \cdot \infty = \infty\).
• The denominator, \(g\left( x \right) = 1 + {x^2}\), approaches \(1 + \infty = \infty\).

Since the limit is of the indeterminate form \(\frac{\infty}{\infty}\), we can apply L'Hopital's Rule. This rule allows us to find the limit of the ratio of two functions by taking the limit of the ratio of their derivatives, provided the original limit is of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).

Applying L'Hopital's Rule (First Time)

Let's find the first derivatives of the numerator and the denominator:

• Derivative of the numerator (\(f\left( x \right) = x \ln\left( x \right)\)):
  Using the product rule \(\frac{d}{{dx}}\left( uv \right) = u'v + uv'\), where \(u=x\) and \(v=\ln(x)\). \[f'\left( x \right) = \frac{d}{{dx}}\left( x \ln\left( x \right) \right) = \left( 1 \cdot \ln\left( x \right) \right) + \left( x \cdot \frac{1}{x} \right) = \ln\left( x \right) + 1\]
• Derivative of the denominator (\(g\left( x \right) = 1 + {x^2}\)):
  \[g'\left( x \right) = \frac{d}{{dx}}\left( 1 + {x^2} \right) = 0 + 2x = 2x\]

Now, applying L'Hopital's Rule, the original limit becomes:

Re-evaluating Limit Form

Let's check the form of this new limit as \(x \to \infty\):

• The numerator, \(\ln\left( x \right) + 1\), approaches \(\infty + 1 = \infty\).
• The denominator, \(2x\), approaches \(\infty\).

Since it's still in the indeterminate form \(\frac{\infty}{\infty}\), we need to apply L'Hopital's Rule a second time.

Applying L'Hopital's Rule (Second Time)

Let \(f_1\left( x \right) = \ln\left( x \right) + 1\) and \(g_1\left( x \right) = 2x\). We find their derivatives:

• Derivative of the new numerator (\(f_1\left( x \right) = \ln\left( x \right) + 1\)):
  \[f_1'\left( x \right) = \frac{d}{{dx}}\left( \ln\left( x \right) + 1 \right) = \frac{1}{x} + 0 = \frac{1}{x}\]
• Derivative of the new denominator (\(g_1\left( x \right) = 2x\)):
  \[g_1'\left( x \right) = \frac{d}{{dx}}\left( 2x \right) = 2\]

Applying L'Hopital's Rule again, the limit becomes:

Final Limit Calculation

Now, simplify the expression and evaluate the limit:

As \(x\) approaches infinity, the denominator \(2x\) also approaches infinity. When the denominator of a fraction becomes infinitely large while the numerator remains a finite non-zero value, the value of the entire fraction approaches zero.

Alternative: Comparison of Growth Rates

Another way to approach this limit is by comparing the growth rates of the functions. As \(x \to \infty\), any positive power of \(x\) grows faster than \(\ln(x)\).

The given expression is \(\frac{{x \ln\left( x \right)}}{{1 + {x^2}}}\).

For very large values of \(x\), the \(1\) in the denominator becomes insignificant, so \(1 + x^2 \approx x^2\).

Thus, the expression can be approximated as:

It is a standard result in calculus that \(\mathop {\lim }\limits_{x \to \infty } \frac{{\ln\left( x \right)}}{x} = 0\). This is because polynomial functions (like \(x\)) grow much faster than logarithmic functions (like \(\ln(x)\)) as \(x\) approaches infinity.

Since the denominator's dominant term (\(x^2\)) grows significantly faster than the numerator's dominant term (\(x \ln(x)\)), the overall fraction approaches 0.

Conclusion

Both methods, L'Hopital's Rule and comparison of growth rates, confirm that the value of the limit is 0.

The final answer is 0.