9. Find the limit. Ukse ľ Hompital's Bule if approprate. It there its a more elementang method, comsider it weing • break, it apart, to check * form is o∞-0, needo to be rewritten into form % or % 9x it I' Hospital's Bule con be used lim and 00 Incx) Incx). 9(0) -9+9 9 1x nx. 9x Clma)-9 (x-1) Inx (x-1) 9x (lnx) -9x +9 = lim Inx (x-1) 961) (In (I)-9 1)+9 = lim rewrite %3D Incx) (x-1) In()(1-1) to) (0) (x-1) Inx メ+」 o indeterminate I'Hoop. again! a 9x (Inx)-9 x-1) Inx(x-1) Inx (x-1) + (x-1) n nx (1x (G) + hox (9)-1 + (x-1)) Inx + (x- 1)(*) 8+ Inx9-9-x+| Inx + 1-/x Inx9-x+ ] x+1 Inx+ 1-/x xx In(I) +!=% lim- = lim = lim = lim In(1)9-1 +1 x+| indeterminate 00
9. Find the limit. Ukse ľ Hompital's Bule if approprate. It there its a more elementang method, comsider it weing • break, it apart, to check * form is o∞-0, needo to be rewritten into form % or % 9x it I' Hospital's Bule con be used lim and 00 Incx) Incx). 9(0) -9+9 9 1x nx. 9x Clma)-9 (x-1) Inx (x-1) 9x (lnx) -9x +9 = lim Inx (x-1) 961) (In (I)-9 1)+9 = lim rewrite %3D Incx) (x-1) In()(1-1) to) (0) (x-1) Inx メ+」 o indeterminate I'Hoop. again! a 9x (Inx)-9 x-1) Inx(x-1) Inx (x-1) + (x-1) n nx (1x (G) + hox (9)-1 + (x-1)) Inx + (x- 1)(*) 8+ Inx9-9-x+| Inx + 1-/x Inx9-x+ ] x+1 Inx+ 1-/x xx In(I) +!=% lim- = lim = lim = lim In(1)9-1 +1 x+| indeterminate 00
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Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
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Please help, I’ve tried and retried. And I just can’t get the answer. The answer is 9/2 but I’m lost. Legit please, explain to me like I’m a baby ??????
![**Title: Calculating Limits Using L'Hôpital's Rule**
**Objective:** Understand the process of finding limits by employing L'Hôpital's Rule when forms are indeterminate.
**Problem Statement:**
Find the limit:
\[ \lim_{x \to 1} \left( \frac{9x}{x-1} - \frac{9}{\ln x} \right) \]
### Steps and Explanation:
1. **Initial Expression Analysis:**
\[
\lim_{x \to 1} \left( \frac{9x}{x-1} - \frac{9}{\ln x} \right)
\]
- Break into separate limits to check if L'Hôpital's Rule is applicable:
\[
\lim_{x \to 1} \frac{9x}{x-1} = \infty \quad \text{and} \quad \lim_{x \to 1} \frac{9}{\ln x} = \infty
\]
- Form is \(\infty - \infty\). Needs rewriting into \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) to apply L'Hôpital's Rule.
2. **Rewriting for L'Hôpital's Rule:**
- Rewriting:
\[
\lim_{x \to 1} \frac{9x \ln x - 9(x-1)}{\ln x (x-1)}
\]
- Simplify numerator:
\[
= \lim_{x \to 1} \frac{9x \ln x - 9x + 9}{\ln x (x-1)}
\]
- Verify form:
\[
= \frac{0}{0} \quad \text{(indeterminate form, apply L'Hôpital)}
\]
3. **Applying L'Hôpital's Rule:**
- Differentiate numerator and denominator:
\[
\lim_{x \to 1} \frac{\frac{d}{dx}(9x \ln x - 9x + 9)}{\frac{d}{dx}(\ln x (x-1))}
\]
- Differentiate numerator:
\[
9 \frac{d}{dx}(x \ln x) + \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8d9f2dd3-4242-4da1-8488-f4a8f54e7aef%2F3fba74b3-06fc-42e8-9f35-0aaac2e43060%2Fh8p5q2k_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Title: Calculating Limits Using L'Hôpital's Rule**
**Objective:** Understand the process of finding limits by employing L'Hôpital's Rule when forms are indeterminate.
**Problem Statement:**
Find the limit:
\[ \lim_{x \to 1} \left( \frac{9x}{x-1} - \frac{9}{\ln x} \right) \]
### Steps and Explanation:
1. **Initial Expression Analysis:**
\[
\lim_{x \to 1} \left( \frac{9x}{x-1} - \frac{9}{\ln x} \right)
\]
- Break into separate limits to check if L'Hôpital's Rule is applicable:
\[
\lim_{x \to 1} \frac{9x}{x-1} = \infty \quad \text{and} \quad \lim_{x \to 1} \frac{9}{\ln x} = \infty
\]
- Form is \(\infty - \infty\). Needs rewriting into \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) to apply L'Hôpital's Rule.
2. **Rewriting for L'Hôpital's Rule:**
- Rewriting:
\[
\lim_{x \to 1} \frac{9x \ln x - 9(x-1)}{\ln x (x-1)}
\]
- Simplify numerator:
\[
= \lim_{x \to 1} \frac{9x \ln x - 9x + 9}{\ln x (x-1)}
\]
- Verify form:
\[
= \frac{0}{0} \quad \text{(indeterminate form, apply L'Hôpital)}
\]
3. **Applying L'Hôpital's Rule:**
- Differentiate numerator and denominator:
\[
\lim_{x \to 1} \frac{\frac{d}{dx}(9x \ln x - 9x + 9)}{\frac{d}{dx}(\ln x (x-1))}
\]
- Differentiate numerator:
\[
9 \frac{d}{dx}(x \ln x) + \
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