Question

1–38 ■ Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t apply, explain why.

19. \(\mathop {\lim }\limits_{x \to 1} \frac{{{x^a} - ax + a - 1}}{{{{(x - 1)}^2}}}\)

Short answer

The value of \(\mathop {\lim }\limits_{x \to 1} \frac{{{x^a} - ax + a - 1}}{{{{(x - 1)}^2}}}\) is \(\frac{{a(a - 1)}}{2}\).

Step-by-step solution

Given information

The given limit is \(\mathop {\lim }\limits_{x \to 1} \frac{{{x^a} - ax + a - 1}}{{{{(x - 1)}^2}}}\).

Concept of L'Hospital's Rule

Suppose\(f\)and\(g\)are differentiable and\({g^\prime }(x) \ne 0\)near\(a\)(except possibly at\(a\)).

Suppose that\(\mathop {\lim }\limits_{x \to a} f(x) = 0\)and\(\mathop {\lim }\limits_{x \to a} g(x) = 0\)or that\(\mathop {\lim }\limits_{x \to a} f(x) = \pm \infty \)and\(\mathop {\lim }\limits_{x \to a} g(x) = \pm \infty \)or in other words, we have an indeterminate form of type\(\frac{0}{0}\)or\(\frac{\infty }{\infty }\).

Then,\(\mathop {\lim }\limits_{x \to a} \frac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to a} \frac{{{f^\prime }(x)}}{{{g^\prime }(x)}}\)if the limit on the right side exists (or is\(\infty \)or\( - \infty \)).

Find the value of the limit

Here \(f(x) = {x^a} - ax + a - 1\) and \(g(x) = {(x - 1)^2}\).

Evaluate the limit of the numerator.

\(\begin{array}{c}\mathop {\lim }\limits_{x \to 1} f(x) = \mathop {\lim }\limits_{x \to 1} {x^a} - ax + a - 1\\ = 1 - a + a - 1\\ = 0\end{array}\)

Evaluate the limit of the denominator.

\(\begin{array}{c}\mathop {\lim }\limits_{x \to 1} g(x) = \mathop {\lim }\limits_{x \to 1} {(x - 1)^2}\\ = {(1 - 1)^2}\\ = 0\end{array}\)

Since \(\mathop {\lim }\limits_{x \to a} f(x) = 0\) and \(\mathop {\lim }\limits_{x \to a} g(x) = 0\), L’Hospital's rule can be applied.

Apply L’Hospital’s rule

Apply L’Hospital's rule and simplify the given limit.

\(\begin{array}{c}\mathop {\lim }\limits_{x \to 1} \frac{{{x^a} - ax + a - 1}}{{{{(x - 1)}^2}}} = \mathop {\lim }\limits_{x \to 1} \frac{{\frac{d}{{dx}}\left( {{x^a} - ax + a - 1} \right)}}{{\frac{d}{{dx}}\left( {{{(x - 1)}^2}} \right)}}\\ = \mathop {\lim }\limits_{x \to 1} \frac{{a{x^{a - 1}} - a}}{{2(x - 1)}}\\ = \frac{{a - a}}{{2(1 - 1)}}\\ = 0\end{array}\)

Since it is \(\frac{0}{0}\) form, L’Hospital's rule can be applied again.

\(\begin{array}{c}\mathop {\lim }\limits_{x \to 1} \frac{{{x^a} - ax + a - 1}}{{{{(x - 1)}^2}}} = \mathop {\lim }\limits_{x \to 1} \frac{{\frac{d}{{dx}}\left( {a{x^{a - 1}} - a} \right)}}{{\frac{d}{{dx}}(2(x - 1))}}\\ = \mathop {\lim }\limits_{x \to 1} \frac{{a(a - 1){x^{a - 2}}}}{2}\\ = \frac{{a(a - 1)}}{2}\end{array}\)

Therefore, the value of \(\mathop {\lim }\limits_{x \to 1} \frac{{{x^a} - ax + a - 1}}{{{{(x - 1)}^2}}}\) is \(\frac{{a(a - 1)}}{2}\).