Question

How do you find the limit of $\left( {2x + 3} \right)\left[ {\ln \left( {{x^2} - x + 1} \right) - \ln \left( {{x^2} + x + 1} \right)} \right]$ as x approaches infinity?

Answer 1

Hint: Given the expression. We have to determine the limit of the expression when x approaches to infinity. First, we will apply the quotient rule of logarithms. Then, determine whether the expression leads to indeterminate state or not. If after applying limits, the result is indeterminate then we will apply L’Hospital’s rule to the expression. Then, we will differentiate the numerator and denominator individually. Then, we will apply the limit to the expression by expanding the expression. Then, determine the coefficient of the highest degree term of the expression and simplify the expression.

Complete step by step solution:
Given the expression, $\left( {2x + 3} \right)\left[ {\ln \left( {{x^2} - x + 1} \right) - \ln \left( {{x^2} + x + 1} \right)} \right]$

We have to find the limit of the expression as x approaches to infinity.

$ \Rightarrow \mathop {\lim }\limits_{x \to \infty } \left( {2x + 3} \right)\left[ {\ln \left( {{x^2} - x + 1} \right) - \ln \left( {{x^2} + x + 1} \right)} \right]$.

Apply the logarithmic rule of difference, $\ln a - \ln b = \ln \dfrac{a}{b}$.

$ \Rightarrow \mathop {\lim }\limits_{x \to \infty } \left( {2x + 3} \right)\left[ {\ln \dfrac{{\left( {{x^2} - x + 1} \right)}}{{\left( {{x^2} + x + 1} \right)}}} \right]$

When the limit is applied to the numerator and denominator of the expression, we get: $ \mathop {\lim }\limits_{x \to \infty } \ln \dfrac{{\left( {{x^2} - x + 1} \right)}}{{\left( {{x^2} + x + 1} \right)}}$

The graph of $\ln x$ tends to infinity when x approaches infinity.

$ \Rightarrow \dfrac{\infty }{\infty }$

Thus, it will lead to an undetermined state.

Now, rewrite the expression as,

$\dfrac{{\ln \left( {{x^2} - x + 1} \right) - \ln \left( {{x^2} + x + 1} \right)}}{{\dfrac{1}{{2x + 3}}}}$

Therefore, we will now differentiate the numerator and denominator of the function individually.

$ \Rightarrow \left[ {\dfrac{{\dfrac{d}{{dx}}\left[ {\ln \left( {{x^2} - x + 1} \right) - \ln \left( {{x^2} + x + 1} \right)} \right]}}{{\dfrac{d}{{dx}}\left( {\dfrac{1}{{2x + 3}}} \right)}}} \right]$

Now, we will apply the chain rule of derivative to differentiate the expression.

$ \Rightarrow \dfrac{{\dfrac{1}{{\left( {{x^2} - x + 1} \right)}} \cdot \dfrac{d}{{dx}}\left( {{x^2} - x + 1} \right) - \dfrac{1}{{\left( {{x^2} + x + 1} \right)}} \cdot \dfrac{d}{{dx}}\left( {{x^2} + x + 1} \right)}}{{\dfrac{d}{{dx}}{{\left( {2x + 3} \right)}^{ - 1}}}}$

$ \Rightarrow \dfrac{{\dfrac{{2x - 1}}{{\left( {{x^2} - x + 1} \right)}} - \dfrac{{2x + 1}}{{\left( {{x^2} + x + 1} \right)}}}}{{\dfrac{{ - 2}}{{{{\left( {2x + 3} \right)}^2}}}}}$

On simplifying the expression further,

$ \Rightarrow \dfrac{{2x - 1}}{{\left( {{x^2} - x + 1} \right)}} - \dfrac{{2x + 1}}{{\left( {{x^2} + x + 1} \right)}} \times \dfrac{{{{\left( {2x + 3} \right)}^2}}}{{ - 2}}$

$ \Rightarrow \dfrac{{\left( {2x - 1} \right)\left( {{x^2} + x + 1} \right) - \left( {2x + 1} \right)\left( {{x^2} - x + 1} \right)}}{{\left( {{x^2} - x + 1} \right)\left( {{x^2} + x + 1} \right)}} \times \dfrac{{{{\left( {2x + 3} \right)}^2}}}{{ - 2}}$

On simplifying the expression further, we get:

$ \Rightarrow \dfrac{{2{x^3} + 2{x^2} + 2x - {x^2} - x - 1 - \left( {2{x^3} - 2{x^2} + 2x + {x^2} - x + 1} \right)}}{{\left( {{x^2} - x + 1} \right)\left( {{x^2} + x + 1} \right)}} \times \dfrac{{{{\left( {2x + 3} \right)}^2}}}{{ - 2}}$

$ \Rightarrow \dfrac{{2{x^3} + 2{x^2} + 2x - {x^2} - x - 1 - 2{x^3} + 2{x^2} - 2x - {x^2} + x - 1}}{{\left( {{x^2} - x + 1} \right)\left( {{x^2} + x + 1} \right)}} \times \dfrac{{{{\left( {2x + 3} \right)}^2}}}{{ - 2}}$

Combine like terms, we get:

$ \Rightarrow \dfrac{{\left( {2{x^2} - 2} \right){{\left( {2x + 3} \right)}^2}}}{{ - 2\left( {{x^2} - x + 1} \right)\left( {{x^2} + x + 1} \right)}}$

Cancel out the common terms.

$ \Rightarrow \dfrac{{2\left( {{x^2} - 1} \right){{\left( {2x + 3} \right)}^2}}}{{ - 2\left( {{x^2} - x + 1} \right)\left( {{x^2} + x + 1} \right)}} = \dfrac{{\left( {1 - {x^2}} \right){{\left( {2x + 3} \right)}^2}}}{{\left( {{x^2} - x + 1} \right)\left( {{x^2} + x + 1} \right)}}$

Apply the limit to the expression.

 $ \Rightarrow \mathop {\lim }\limits_{x \to \infty } \dfrac{{\left( {1 - {x^2}} \right){{\left( {2x + 3} \right)}^2}}}{{\left( {{x^2} - x + 1} \right)\left( {{x^2} + x + 1} \right)}}$

Substitute $x = \infty $to the numerator and denominator.

$ \Rightarrow \mathop {\lim }\limits_{x \to \infty } \dfrac{{\left( {1 - {x^2}} \right)\left( {4{x^2} + 12x + 9} \right)}}{{\left( {{x^2} - x + 1} \right)\left( {{x^2} + x + 1} \right)}}$

Expand the polynomials, we get:

\[ \Rightarrow \dfrac{{ - 4{x^4} + {\text{ terms of lesser degree}}}}{{{x^4} + {\text{ terms of lesser degree}}}}\]

So, the limit will be

\[ \Rightarrow \dfrac{{ - 4{x^4}}}{{{x^4}}} = - 4\]

Hence the limit of $\left( {2x + 3} \right)\left[ {\ln \left( {{x^2} - x + 1} \right) - \ln \left( {{x^2} + x + 1} \right)} \right]$ is \[ - 4\] as x approaches infinity.

Note: Please note that in such types of questions, we can get indeterminate form. In such cases we have to apply the L’Hospital’s rule to the function, which is defined as \[\mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to a} \dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}\].