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SN1F24 Quiz8: Problem 1 (1 point)
Evaluate the limit, using L'Hôpital's Rule. Enter INF for $\infty,-$ INF for $-\infty$ , or DNE if the limit does not exist, but is neither $\infty$ nor $-\infty$ . $\lim _{x \rightarrow 0^{+}} 13 x \ln x=$ $\square$ Preview My Answers Submit Answers
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Studdy Solution

STEP 1

1. We are asked to evaluate the limit $\lim_{x \rightarrow 0^{+}} 13x \ln x$ .
2. We will use L'Hôpital's Rule, which applies to indeterminate forms like $\frac{0}{0}$ or $\frac{\infty}{\infty}$ .

STEP 2

1. Identify the indeterminate form.
2. Apply L'Hôpital's Rule.
3. Differentiate the numerator and the denominator.
4. Evaluate the limit after applying L'Hôpital's Rule.
5. Determine if the limit exists or if it is $\infty$ , $-\infty$ , or DNE.

STEP 3

Identify the indeterminate form:
The expression $13x \ln x$ can be rewritten as $\frac{\ln x}{1/(13x)}$ .
As $x \rightarrow 0^{+}$ , $\ln x \rightarrow -\infty$ and $1/(13x) \rightarrow \infty$ , which is an indeterminate form of $-\infty/\infty$ .

STEP 4

Apply L'Hôpital's Rule:
Since we have an indeterminate form of $-\infty/\infty$ , we can apply L'Hôpital's Rule, which involves differentiating the numerator and the denominator.

STEP 5

Differentiate the numerator and the denominator:
The numerator $\ln x$ differentiates to $\frac{1}{x}$ .
The denominator $1/(13x)$ differentiates to $-\frac{1}{13x^2}$ .

STEP 6

Evaluate the limit after applying L'Hôpital's Rule:
The limit becomes:
$\lim_{x \rightarrow 0^{+}} \frac{\frac{1}{x}}{-\frac{1}{13x^2}} = \lim_{x \rightarrow 0^{+}} -13x$
As $x \rightarrow 0^{+}$ , $-13x \rightarrow 0$ .

STEP 7

Determine if the limit exists or if it is $\infty$ , $-\infty$ , or DNE:
The limit is $0$ , which is a finite value, so it exists.
The value of the limit is:
$\boxed{0}$

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Studdy Solution

STEP 1

1. We are asked to evaluate the limit $\lim_{x \rightarrow 0^{+}} 13x \ln x$ .
2. We will use L'Hôpital's Rule, which applies to indeterminate forms like $\frac{0}{0}$ or $\frac{\infty}{\infty}$ .

STEP 2

1. Identify the indeterminate form.
2. Apply L'Hôpital's Rule.
3. Differentiate the numerator and the denominator.
4. Evaluate the limit after applying L'Hôpital's Rule.
5. Determine if the limit exists or if it is $\infty$ , $-\infty$ , or DNE.

STEP 3

Identify the indeterminate form:
The expression $13x \ln x$ can be rewritten as $\frac{\ln x}{1/(13x)}$ .
As $x \rightarrow 0^{+}$ , $\ln x \rightarrow -\infty$ and $1/(13x) \rightarrow \infty$ , which is an indeterminate form of $-\infty/\infty$ .

STEP 4

Apply L'Hôpital's Rule:
Since we have an indeterminate form of $-\infty/\infty$ , we can apply L'Hôpital's Rule, which involves differentiating the numerator and the denominator.

STEP 5

Differentiate the numerator and the denominator:
The numerator $\ln x$ differentiates to $\frac{1}{x}$ .
The denominator $1/(13x)$ differentiates to $-\frac{1}{13x^2}$ .

STEP 6

Evaluate the limit after applying L'Hôpital's Rule:
The limit becomes:
$\lim_{x \rightarrow 0^{+}} \frac{\frac{1}{x}}{-\frac{1}{13x^2}} = \lim_{x \rightarrow 0^{+}} -13x$
As $x \rightarrow 0^{+}$ , $-13x \rightarrow 0$ .

STEP 7

Determine if the limit exists or if it is $\infty$ , $-\infty$ , or DNE:
The limit is $0$ , which is a finite value, so it exists.
The value of the limit is:
$\boxed{0}$

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Studdy Solution

STEP 1

1. We are asked to evaluate the limit lim⁡x→0+13xln⁡x\lim_{x \rightarrow 0^{+}} 13x \ln xlimx→0+​13xlnx.2. We will use L'Hôpital's Rule, which applies to indeterminate forms like 00\frac{0}{0}00​ or ∞∞\frac{\infty}{\infty}∞∞​.

STEP 2

1. Identify the indeterminate form.2. Apply L'Hôpital's Rule.3. Differentiate the numerator and the denominator.4. Evaluate the limit after applying L'Hôpital's Rule.5. Determine if the limit exists or if it is ∞\infty∞, −∞-\infty−∞, or DNE.

STEP 3

STEP 4

Apply L'Hôpital's Rule:Since we have an indeterminate form of −∞/∞-\infty/\infty−∞/∞, we can apply L'Hôpital's Rule, which involves differentiating the numerator and the denominator.

STEP 5

Differentiate the numerator and the denominator:The numerator ln⁡x\ln xlnx differentiates to 1x\frac{1}{x}x1​.The denominator 1/(13x)1/(13x)1/(13x) differentiates to −113x2-\frac{1}{13x^2}−13x21​.

STEP 6

STEP 7

Determine if the limit exists or if it is ∞\infty∞, −∞-\infty−∞, or DNE:The limit is 000, which is a finite value, so it exists. The value of the limit is:0 \boxed{0} 0​

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1. We are asked to evaluate the limit $\lim_{x \rightarrow 0^{+}} 13x \ln x$ .
2. We will use L'Hôpital's Rule, which applies to indeterminate forms like $\frac{0}{0}$ or $\frac{\infty}{\infty}$ .

1. Identify the indeterminate form.
2. Apply L'Hôpital's Rule.
3. Differentiate the numerator and the denominator.
4. Evaluate the limit after applying L'Hôpital's Rule.
5. Determine if the limit exists or if it is $\infty$ , $-\infty$ , or DNE.

Apply L'Hôpital's Rule:
Since we have an indeterminate form of $-\infty/\infty$ , we can apply L'Hôpital's Rule, which involves differentiating the numerator and the denominator.

Differentiate the numerator and the denominator:
The numerator $\ln x$ differentiates to $\frac{1}{x}$ .
The denominator $1/(13x)$ differentiates to $-\frac{1}{13x^2}$ .

Determine if the limit exists or if it is $\infty$ , $-\infty$ , or DNE:
The limit is $0$ , which is a finite value, so it exists.
The value of the limit is:
$\boxed{0}$