Question

35-38: Use continuity to evaluate the limit.

37. \(\mathop {\lim }\limits_{x \to 1} \ln \left( {\frac{{5 - {x^2}}}{{1 + x}}} \right)\)

Short answer

The solution of the limit is \(\ln 2\).

Step-by-step solution

Theorems of continuity

When the functions say\(f\)and\(g\)arecontinuous at some number. Then, the following functions are also continuous:

1. \(f + g\)
2. \(f - g\)
3. \(cf\)
4. \(fg\)
5. \(\frac{f}{g},\,\,{\mathop{\rm if}\nolimits} \,\,g\left( a \right) \ne 0\)

Eachpolynomialiscontinuouson\(\mathbb{R} = \left( { - \infty ,\infty } \right)\). Anyrational functionis continuous in its domain.

When \(f\), and \(g\) are continuous at some number, then thecomposite function\(f \circ g\)is\(\left( {f \circ g} \right)\left( x \right) = f\left( {g\left( x \right)} \right)\)also continuous.

Use continuity to evaluate the limit

It is observed that the function\(f\left( x \right) = \ln \left( {\frac{{5 - {x^2}}}{{1 + x}}} \right)\)is the composite of a logarithm function and a rational function.

Therefore, the function\(f\left( x \right)\)is continuous throughout its domain.

We should have\(\frac{{5 - {x^2}}}{{1 + x}} > 0\)for the domain of the function\(f\). The sign of the function's numerator and denominator should match.

Therefore, the domain of the function is\(\left( { - \infty , - \sqrt 5 } \right) \cup \left( { - 1,\sqrt 5 } \right)\).

The function\(f\)is continuous at 1.

Evaluate the limit as shown below:

\(\begin{aligned}\mathop {\lim }\limits_{x \to 1} f\left( x \right) &= f\left( 1 \right)\\ &= \ln \frac{{5 - 1}}{{1 + 1}}\\ &= \ln 2\end{aligned}\)

Hence, the solution of the limit is \(\ln 2\).