Math / Calculus

Question(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 $-\infty$ . $\lim _{x \rightarrow 0^{+}}(1+3 x)^{3 / x}=$ $\square$
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Studdy Solution

STEP 1

1. We are given the limit: $\lim_{x \rightarrow 0^{+}}(1+3x)^{3/x}$ .
2. We need to evaluate this limit using L'Hôpital's Rule.
3. L'Hôpital's Rule can be applied to limits of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ .

STEP 2

1. Recognize the form of the limit and transform it to a suitable form for L'Hôpital's Rule.
2. Take the natural logarithm to simplify the expression.
3. Differentiate the numerator and the denominator.
4. Apply L'Hôpital's Rule.
5. Evaluate the limit and exponentiate to find the original limit.

STEP 3

Recognize that the given limit is of the form $(1+3x)^{3/x}$ . To use L'Hôpital's Rule, we first take the natural logarithm:
Let $y = \lim_{x \rightarrow 0^{+}}(1+3x)^{3/x}$ .
Take the natural logarithm of both sides:
$\ln y = \lim_{x \rightarrow 0^{+}} \frac{3}{x} \ln(1+3x)$
This transforms the problem into finding the limit of a quotient.

STEP 4

Now, we have:
$\ln y = \lim_{x \rightarrow 0^{+}} \frac{3 \ln(1+3x)}{x}$
This is an indeterminate form $\frac{0}{0}$ as $x \rightarrow 0^{+}$ .

STEP 5

Differentiate the numerator and the denominator:
Numerator: $\frac{d}{dx}[3 \ln(1+3x)] = \frac{3 \cdot 3}{1+3x} = \frac{9}{1+3x}$
Denominator: $\frac{d}{dx}[x] = 1$

STEP 6

Apply L'Hôpital's Rule:
$\ln y = \lim_{x \rightarrow 0^{+}} \frac{9}{1+3x}$
Evaluate the limit as $x \rightarrow 0^{+}$ :
$\ln y = \frac{9}{1+0} = 9$

STEP 7

Exponentiate to solve for $y$ :
$y = e^{9}$
The original limit is:
$\lim_{x \rightarrow 0^{+}}(1+3x)^{3/x} = e^{9}$
The value of the limit is:
$\boxed{e^{9}}$

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Studdy Solution

STEP 1

1. We are given the limit: $\lim_{x \rightarrow 0^{+}}(1+3x)^{3/x}$ .
2. We need to evaluate this limit using L'Hôpital's Rule.
3. L'Hôpital's Rule can be applied to limits of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ .

STEP 2

1. Recognize the form of the limit and transform it to a suitable form for L'Hôpital's Rule.
2. Take the natural logarithm to simplify the expression.
3. Differentiate the numerator and the denominator.
4. Apply L'Hôpital's Rule.
5. Evaluate the limit and exponentiate to find the original limit.

STEP 3

STEP 4

Now, we have:
$\ln y = \lim_{x \rightarrow 0^{+}} \frac{3 \ln(1+3x)}{x}$
This is an indeterminate form $\frac{0}{0}$ as $x \rightarrow 0^{+}$ .

STEP 5

Differentiate the numerator and the denominator:
Numerator: $\frac{d}{dx}[3 \ln(1+3x)] = \frac{3 \cdot 3}{1+3x} = \frac{9}{1+3x}$
Denominator: $\frac{d}{dx}[x] = 1$

STEP 6

Apply L'Hôpital's Rule:
$\ln y = \lim_{x \rightarrow 0^{+}} \frac{9}{1+3x}$
Evaluate the limit as $x \rightarrow 0^{+}$ :
$\ln y = \frac{9}{1+0} = 9$

STEP 7

Exponentiate to solve for $y$ :
$y = e^{9}$
The original limit is:
$\lim_{x \rightarrow 0^{+}}(1+3x)^{3/x} = e^{9}$
The value of the limit is:
$\boxed{e^{9}}$

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Studdy Solution

STEP 1

1. We are given the limit: lim⁡x→0+(1+3x)3/x\lim_{x \rightarrow 0^{+}}(1+3x)^{3/x}limx→0+​(1+3x)3/x.2. We need to evaluate this limit using L'Hôpital's Rule.3. L'Hôpital's Rule can be applied to limits of the form 00\frac{0}{0}00​ or ∞∞\frac{\infty}{\infty}∞∞​.

STEP 2

1. Recognize the form of the limit and transform it to a suitable form for L'Hôpital's Rule.2. Take the natural logarithm to simplify the expression.3. Differentiate the numerator and the denominator.4. Apply L'Hôpital's Rule.5. Evaluate the limit and exponentiate to find the original limit.

STEP 3

STEP 4

Now, we have:ln⁡y=lim⁡x→0+3ln⁡(1+3x)x \ln y = \lim_{x \rightarrow 0^{+}} \frac{3 \ln(1+3x)}{x} lny=x→0+lim​x3ln(1+3x)​This is an indeterminate form 00\frac{0}{0}00​ as x→0+x \rightarrow 0^{+}x→0+.

STEP 5

Differentiate the numerator and the denominator:Numerator: ddx[3ln⁡(1+3x)]=3⋅31+3x=91+3x \frac{d}{dx}[3 \ln(1+3x)] = \frac{3 \cdot 3}{1+3x} = \frac{9}{1+3x} dxd​[3ln(1+3x)]=1+3x3⋅3​=1+3x9​Denominator: ddx[x]=1 \frac{d}{dx}[x] = 1 dxd​[x]=1

STEP 6

Apply L'Hôpital's Rule:ln⁡y=lim⁡x→0+91+3x \ln y = \lim_{x \rightarrow 0^{+}} \frac{9}{1+3x} lny=x→0+lim​1+3x9​Evaluate the limit as x→0+x \rightarrow 0^{+}x→0+:ln⁡y=91+0=9 \ln y = \frac{9}{1+0} = 9 lny=1+09​=9

STEP 7

Exponentiate to solve for yyy:y=e9 y = e^{9} y=e9The original limit is:lim⁡x→0+(1+3x)3/x=e9 \lim_{x \rightarrow 0^{+}}(1+3x)^{3/x} = e^{9} x→0+lim​(1+3x)3/x=e9 The value of the limit is:e9 \boxed{e^{9}} e9​

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. We are given the limit: $\lim_{x \rightarrow 0^{+}}(1+3x)^{3/x}$ .
2. We need to evaluate this limit using L'Hôpital's Rule.
3. L'Hôpital's Rule can be applied to limits of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ .

1. Recognize the form of the limit and transform it to a suitable form for L'Hôpital's Rule.
2. Take the natural logarithm to simplify the expression.
3. Differentiate the numerator and the denominator.
4. Apply L'Hôpital's Rule.
5. Evaluate the limit and exponentiate to find the original limit.

Now, we have:
$\ln y = \lim_{x \rightarrow 0^{+}} \frac{3 \ln(1+3x)}{x}$
This is an indeterminate form $\frac{0}{0}$ as $x \rightarrow 0^{+}$ .

Differentiate the numerator and the denominator:
Numerator: $\frac{d}{dx}[3 \ln(1+3x)] = \frac{3 \cdot 3}{1+3x} = \frac{9}{1+3x}$
Denominator: $\frac{d}{dx}[x] = 1$

Apply L'Hôpital's Rule:
$\ln y = \lim_{x \rightarrow 0^{+}} \frac{9}{1+3x}$
Evaluate the limit as $x \rightarrow 0^{+}$ :
$\ln y = \frac{9}{1+0} = 9$

Exponentiate to solve for $y$ :
$y = e^{9}$
The original limit is:
$\lim_{x \rightarrow 0^{+}}(1+3x)^{3/x} = e^{9}$
The value of the limit is:
$\boxed{e^{9}}$