Calculus Examples

$lim_{x \to 1} (x - 1) \ln (x - 1)$

Step 1

Change the two-sided limit into a right sided limit.

$lim_{x \to 1^{+}} (x - 1) \ln (x - 1)$

Step 2

Rewrite $(x - 1) \ln (x - 1)$ as $\frac{x - 1}{\frac{1}{\ln (x - 1)}}$ .

$lim_{x \to 1^{+}} \frac{x - 1}{\frac{1}{\ln (x - 1)}}$

Step 3

Apply L'Hospital's rule.

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Step 3.1

Evaluate the limit of the numerator and the limit of the denominator.

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Step 3.1.1

Take the limit of the numerator and the limit of the denominator.

$\frac{lim_{x \to 1^{+}} x - 1}{lim_{x \to 1^{+}} \frac{1}{\ln (x - 1)}}$

Step 3.1.2

Evaluate the limit of the numerator.

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Step 3.1.2.1

Evaluate the limit.

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Step 3.1.2.1.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $1$ .

$\frac{lim_{x \to 1^{+}} x - lim_{x \to 1^{+}} 1}{lim_{x \to 1^{+}} \frac{1}{\ln (x - 1)}}$

Step 3.1.2.1.2

Evaluate the limit of $1$ which is constant as $x$ approaches $1$ .

$\frac{lim_{x \to 1^{+}} x - 1 \cdot 1}{lim_{x \to 1^{+}} \frac{1}{\ln (x - 1)}}$

Step 3.1.2.2

Evaluate the limit of $x$ by plugging in $1$ for $x$ .

$\frac{1 - 1 \cdot 1}{lim_{x \to 1^{+}} \frac{1}{\ln (x - 1)}}$

Step 3.1.2.3

Simplify the answer.

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Step 3.1.2.3.1

Multiply $- 1$ by $1$ .

$\frac{1 - 1}{lim_{x \to 1^{+}} \frac{1}{\ln (x - 1)}}$

Step 3.1.2.3.2

Subtract $1$ from $1$ .

$\frac{0}{lim_{x \to 1^{+}} \frac{1}{\ln (x - 1)}}$

Step 3.1.3

Since its numerator approaches a real number while its denominator is unbounded, the fraction $\frac{1}{\ln (x - 1)}$ approaches $0$ .

$\frac{0}{0}$

Step 3.1.4

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{0}{0}$

Step 3.2

Since $\frac{0}{0}$ is of indeterminate form, apply L'Hospital's Rule. L'Hospital's Rule states that the limit of a quotient of functions is equal to the limit of the quotient of their derivatives.

$lim_{x \to 1^{+}} \frac{x - 1}{\frac{1}{\ln (x - 1)}} = lim_{x \to 1^{+}} \frac{\frac{d}{d x} [x - 1]}{\frac{d}{d x} [\frac{1}{\ln (x - 1)}]}$

Step 3.3

Find the derivative of the numerator and denominator.

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Step 3.3.1

Differentiate the numerator and denominator.

$lim_{x \to 1^{+}} \frac{\frac{d}{d x} [x - 1]}{\frac{d}{d x} [\frac{1}{\ln (x - 1)}]}$

Step 3.3.2

By the Sum Rule, the derivative of $x - 1$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 1]$ .

$lim_{x \to 1^{+}} \frac{\frac{d}{d x} [x] + \frac{d}{d x} [- 1]}{\frac{d}{d x} [\frac{1}{\ln (x - 1)}]}$

Step 3.3.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$lim_{x \to 1^{+}} \frac{1 + \frac{d}{d x} [- 1]}{\frac{d}{d x} [\frac{1}{\ln (x - 1)}]}$

Step 3.3.4

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$lim_{x \to 1^{+}} \frac{1 + 0}{\frac{d}{d x} [\frac{1}{\ln (x - 1)}]}$

Step 3.3.5

Add $1$ and $0$ .

$lim_{x \to 1^{+}} \frac{1}{\frac{d}{d x} [\frac{1}{\ln (x - 1)}]}$

Step 3.3.6

Rewrite $\frac{1}{\ln (x - 1)}$ as $\ln^{- 1} (x - 1)$ .

$lim_{x \to 1^{+}} \frac{1}{\frac{d}{d x} [\ln^{- 1} (x - 1)]}$

Step 3.3.7

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- 1}$ and $g (x) = \ln (x - 1)$ .

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Step 3.3.7.1

To apply the Chain Rule, set $u_{1}$ as $\ln (x - 1)$ .

$lim_{x \to 1^{+}} \frac{1}{\frac{d}{d u_{1}} [{u_{1}}^{- 1}] \frac{d}{d x} [\ln (x - 1)]}$

Step 3.3.7.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{1}} [{u_{1}}^{n}]$ is $n {u_{1}}^{n - 1}$ where $n = - 1$ .

$lim_{x \to 1^{+}} \frac{1}{- {u_{1}}^{- 2} \frac{d}{d x} [\ln (x - 1)]}$

Step 3.3.7.3

Replace all occurrences of $u_{1}$ with $\ln (x - 1)$ .

$lim_{x \to 1^{+}} \frac{1}{- \ln^{- 2} (x - 1) \frac{d}{d x} [\ln (x - 1)]}$

Step 3.3.8

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = x - 1$ .

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Step 3.3.8.1

To apply the Chain Rule, set $u_{2}$ as $x - 1$ .

$lim_{x \to 1^{+}} \frac{1}{- \ln^{- 2} (x - 1) (\frac{d}{d u_{2}} [\ln (u_{2})] \frac{d}{d x} [x - 1])}$

Step 3.3.8.2

The derivative of $\ln (u_{2})$ with respect to $u_{2}$ is $\frac{1}{u_{2}}$ .

$lim_{x \to 1^{+}} \frac{1}{- \ln^{- 2} (x - 1) (\frac{1}{u_{2}} \frac{d}{d x} [x - 1])}$

Step 3.3.8.3

Replace all occurrences of $u_{2}$ with $x - 1$ .

$lim_{x \to 1^{+}} \frac{1}{- \ln^{- 2} (x - 1) (\frac{1}{x - 1} \frac{d}{d x} [x - 1])}$

Step 3.3.9

Combine $\frac{1}{x - 1}$ and $\ln^{- 2} (x - 1)$ .

$lim_{x \to 1^{+}} \frac{1}{- \frac{\ln^{- 2} (x - 1)}{x - 1} \frac{d}{d x} [x - 1]}$

Step 3.3.10

Move $\ln^{- 2} (x - 1)$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$lim_{x \to 1^{+}} \frac{1}{- \frac{1}{(x - 1) \ln^{2} (x - 1)} \frac{d}{d x} [x - 1]}$

Step 3.3.11

By the Sum Rule, the derivative of $x - 1$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 1]$ .

$lim_{x \to 1^{+}} \frac{1}{- \frac{1}{(x - 1) \ln^{2} (x - 1)} (\frac{d}{d x} [x] + \frac{d}{d x} [- 1])}$

Step 3.3.12

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$lim_{x \to 1^{+}} \frac{1}{- \frac{1}{(x - 1) \ln^{2} (x - 1)} (1 + \frac{d}{d x} [- 1])}$

Step 3.3.13

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$lim_{x \to 1^{+}} \frac{1}{- \frac{1}{(x - 1) \ln^{2} (x - 1)} (1 + 0)}$

Step 3.3.14

Add $1$ and $0$ .

$lim_{x \to 1^{+}} \frac{1}{- \frac{1}{(x - 1) \ln^{2} (x - 1)} \cdot 1}$

Step 3.3.15

Multiply $- 1$ by $1$ .

$lim_{x \to 1^{+}} \frac{1}{- \frac{1}{(x - 1) \ln^{2} (x - 1)}}$

Step 3.3.16

Reorder factors in $- \frac{1}{(x - 1) \ln^{2} (x - 1)}$ .

$lim_{x \to 1^{+}} \frac{1}{- \frac{1}{\ln^{2} (x - 1) (x - 1)}}$

Step 3.4

Multiply the numerator by the reciprocal of the denominator.

$lim_{x \to 1^{+}} 1 (- (\ln^{2} (x - 1) (x - 1)))$

Step 3.5

Multiply $- 1$ by $1$ .

$lim_{x \to 1^{+}} - (\ln^{2} (x - 1) (x - 1))$

Step 3.6

Apply the distributive property.

$lim_{x \to 1^{+}} - (\ln^{2} (x - 1) x + \ln^{2} (x - 1) \cdot - 1)$

Step 3.7

Move $- 1$ to the left of $\ln^{2} (x - 1)$ .

$lim_{x \to 1^{+}} - (\ln^{2} (x - 1) x - 1 \cdot \ln^{2} (x - 1))$

Step 3.8

Rewrite $- 1 \ln^{2} (x - 1)$ as $- \ln^{2} (x - 1)$ .

$lim_{x \to 1^{+}} - (\ln^{2} (x - 1) x - \ln^{2} (x - 1))$

Step 3.9

Apply the distributive property.

$lim_{x \to 1^{+}} - (\ln^{2} (x - 1) x) - - \ln^{2} (x - 1)$

Step 3.10

Multiply $- - \ln^{2} (x - 1)$ .

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Step 3.10.1

Multiply $- 1$ by $- 1$ .

$lim_{x \to 1^{+}} - \ln^{2} (x - 1) x + 1 \ln^{2} (x - 1)$

Step 3.10.2

Multiply $\ln^{2} (x - 1)$ by $1$ .

$lim_{x \to 1^{+}} - \ln^{2} (x - 1) x + \ln^{2} (x - 1)$

Step 3.11

Reorder factors in $- \ln^{2} (x - 1) x + \ln^{2} (x - 1)$ .

$lim_{x \to 1^{+}} - x \ln^{2} (x - 1) + \ln^{2} (x - 1)$

Step 4

Make a table to show the behavior of the function $- x \ln^{2} (x - 1) + \ln^{2} (x - 1)$ as $x$ approaches $1$ from the right.

$\begin{array}{c} x & - x \ln^{2} (x - 1) + \ln^{2} (x - 1) \\ 1.1 & - 0.53018981 \\ 1.01 & - 0.21207592 \\ 1.001 & - 0.04771708 \\ 1.0001 & - 0.00848303 \\ 1.00001 & - 0.00132547 \\ 1.000001 & - 0.00019086 \\ 1.0000001 & - 0.00002597 \end{array}$

Step 5

As the $x$ values approach $1$ , the function values approach $0$ . Thus, the limit of $- x \ln^{2} (x - 1) + \ln^{2} (x - 1)$ as $x$ approaches $1$ from the right is $0$ .

$0$