Limits, Sequences, Series: Learning Objective 3

Direct substitution results in the indeterminate form $\frac{0}{0}$. For the first method, try factoring the numerator and denominator. For the second method, apply L'Hôpital's Rule by taking the derivative of the numerator and the denominator.

$\frac{1}{2}$

Method 1: Without L'Hôpital's Rule (Factoring)
We factor the numerator and the denominator: $$ \lim_{x \to 1} \frac{x^2-x}{x^2-1} = \lim_{x \to 1} \frac{x(x-1)}{(x-1)(x+1)} $$ Since $x \to 1$, $x \neq 1$, so we can cancel the $(x-1)$ terms: $$ = \lim_{x \to 1} \frac{x}{x+1} $$ Now, using direct substitution: $$ = \frac{1}{1+1} = \frac{1}{2} $$

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Method 2: Using L'Hôpital's Rule
We have the indeterminate form $\frac{0}{0}$. We take the derivative of the numerator, $f'(x) = 2x-1$, and the denominator, $g'(x) = 2x$. $$ \lim_{x \to 1} \frac{x^2-x}{x^2-1} \overset{L'H}{=} \lim_{x \to 1} \frac{2x-1}{2x} $$ Now, using direct substitution: $$ = \frac{2(1)-1}{2(1)} = \frac{1}{2} $$

Direct substitution gives $\frac{\ln(1)}{1-1} = \frac{0}{0}$. This is an indeterminate form, so L'Hôpital's Rule can be applied.

$1$

1. Check for Indeterminate Form: As $x \to 1$, the numerator $\ln(x) \to \ln(1) = 0$ and the denominator $x-1 \to 1-1=0$. We have the form $\frac{0}{0}$.

2. Apply L'Hôpital's Rule: We differentiate the numerator and the denominator:

• $f'(x) = \frac{d}{dx}(\ln x) = \frac{1}{x}$
• $g'(x) = \frac{d}{dx}(x-1) = 1$

$$ \lim_{x \to 1} \frac{\ln(x)}{x-1} \overset{L'H}{=} \lim_{x \to 1} \frac{1/x}{1} $$ 3. Evaluate the New Limit: $$ = \lim_{x \to 1} \frac{1}{x} = \frac{1}{1} = 1 $$

The limit is of the indeterminate form $\frac{\infty}{\infty}$. You will need to apply L'Hôpital's Rule more than once.

$\infty$

1. Check for Indeterminate Form: As $x \to \infty$, $e^x \to \infty$ and $x^2 \to \infty$. We have the form $\frac{\infty}{\infty}$.

2. Apply L'Hôpital's Rule (First Time): $$ \lim_{x \to \infty} \frac{e^x}{x^2} \overset{L'H}{=} \lim_{x \to \infty} \frac{\frac{d}{dx}(e^x)}{\frac{d}{dx}(x^2)} = \lim_{x \to \infty} \frac{e^x}{2x} $$ 3. Check Again: The new limit is still of the form $\frac{\infty}{\infty}$. We must apply the rule again.

4. Apply L'Hôpital's Rule (Second Time): $$ \lim_{x \to \infty} \frac{e^x}{2x} \overset{L'H}{=} \lim_{x \to \infty} \frac{\frac{d}{dx}(e^x)}{\frac{d}{dx}(2x)} = \lim_{x \to \infty} \frac{e^x}{2} $$ 5. Evaluate the Final Limit: As $x \to \infty$, the numerator $e^x$ grows without bound while the denominator is a constant. $$ \lim_{x \to \infty} \frac{e^x}{2} = \infty $$

This results in the indeterminate form $\frac{0}{0}$. You can solve this either by factoring the numerator or by applying L'Hôpital's Rule.

$6$

Method 1: Factoring $$ \lim_{x \to 4} \frac{x^2-2x-8}{x-4} = \lim_{x \to 4} \frac{(x-4)(x+2)}{x-4} $$ Cancel the $(x-4)$ terms: $$ = \lim_{x \to 4} (x+2) = 4+2=6 $$

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Method 2: L'Hôpital's Rule
The limit is of the form $\frac{4^2-2(4)-8}{4-4} = \frac{16-8-8}{0} = \frac{0}{0}$. $$ \lim_{x \to 4} \frac{x^2-2x-8}{x-4} \overset{L'H}{=} \lim_{x \to 4} \frac{\frac{d}{dx}(x^2-2x-8)}{\frac{d}{dx}(x-4)} = \lim_{x \to 4} \frac{2x-2}{1} $$ Evaluate the new limit: $$ = 2(4)-2 = 6 $$

First, try direct substitution. L'Hôpital's rule only applies to indeterminate forms like $\frac{0}{0}$ or $\frac{\infty}{\infty}$. Check if the conditions are met before applying the rule.

$0$. It is important to check that the conditions to use L'Hôpital's rule are satisfied before applying it.

1. Evaluate the limit of the numerator and denominator by direct substitution:

• Numerator: $\lim_{x \to \pi^{-}} \sin(x) = \sin(\pi) = 0$
• Denominator: $\lim_{x \to \pi^{-}} (1-\cos(x)) = 1 - \cos(\pi) = 1 - (-1) = 2$

2. Determine the limit: Since the limit of the denominator is not 0, this is not an indeterminate form. We can evaluate the limit directly. $$ \lim_{x \to \pi^{-}} \frac{\sin(x)}{1-\cos(x)} = \frac{0}{2} = 0 $$ L'Hôpital's rule does not apply here because the limit is not of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$. Applying it would lead to an incorrect result: $$ \lim_{x \to \pi^{-}} \frac{\cos(x)}{\sin(x)} \to \frac{-1}{0^+} \to -\infty \quad (\text{Incorrect}) $$

This is an indeterminate product of the form $0 \cdot (-\infty)$. To use L'Hôpital's rule, rewrite the expression as a fraction, such as $\frac{\ln(x)}{1/x}$.

$0$

1. Identify Indeterminate Form: As $x \to 0^+$, we have $x \to 0$ and $\ln(x) \to -\infty$. This is the indeterminate form $0 \cdot (-\infty)$.

2. Rewrite as a Quotient: We can rewrite $x\ln(x)$ as $\frac{\ln(x)}{1/x}$. Now as $x \to 0^+$, the numerator $\ln(x) \to -\infty$ and the denominator $1/x \to \infty$. This is the indeterminate form $\frac{-\infty}{\infty}$, so L'Hôpital's Rule applies.

3. Apply L'Hôpital's Rule: $$ \lim_{x \to 0^+} \frac{\ln(x)}{1/x} \overset{L'H}{=} \lim_{x \to 0^+} \frac{\frac{d}{dx}(\ln x)}{\frac{d}{dx}(x^{-1})} = \lim_{x \to 0^+} \frac{1/x}{-1x^{-2}} = \lim_{x \to 0^+} \frac{1/x}{-1/x^2} $$ 4. Simplify and Evaluate: $$ = \lim_{x \to 0^+} \left( \frac{1}{x} \cdot -x^2 \right) = \lim_{x \to 0^+} (-x) = 0 $$

This is an indeterminate power of the form $0^0$. Use logarithms to handle this. Let $y = x^x$, then take the natural logarithm of both sides: $\ln(y) = x\ln(x)$. Find the limit of $\ln(y)$ first, then exponentiate to find the limit of $y$.

$1$

Let $L = \lim_{x \to 0^+} x^x$. This is of the form $0^0$.

1. Use Logarithms: Let $y = x^x$. Then $\ln(y) = \ln(x^x) = x\ln(x)$.

2. Find the Limit of the Logarithm: We want to find $\lim_{x \to 0^+} \ln(y) = \lim_{x \to 0^+} x\ln(x)$. From the previous problem, we know this limit is 0. $$ \lim_{x \to 0^+} \ln(y) = 0 $$ 3. Exponentiate to Find the Original Limit: Since we found the limit of $\ln(y)$, we can now find the limit of $y$. $$ L = \lim_{x \to 0^+} y = \lim_{x \to 0^+} e^{\ln(y)} $$ Because the exponential function is continuous, we can bring the limit inside: $$ L = e^{\lim_{x \to 0^+} \ln(y)} = e^0 = 1 $$

Direct substitution gives the indeterminate form $\frac{1-1}{0} = \frac{0}{0}$. Apply L'Hôpital's Rule. Recall that the derivative of $a^t$ is $a^t \ln(a)$.

$\ln(\frac{8}{5})$

1. Check for Indeterminate Form: As $t \to 0$, we have $8^0 - 5^0 = 1 - 1 = 0$ in the numerator and $0$ in the denominator. This is the form $\frac{0}{0}$.

2. Apply L'Hôpital's Rule: $$ \lim_{t \to 0} \frac{8^t-5^t}{t} \overset{L'H}{=} \lim_{t \to 0} \frac{\frac{d}{dt}(8^t-5^t)}{\frac{d}{dt}(t)} = \lim_{t \to 0} \frac{8^t\ln(8) - 5^t\ln(5)}{1} $$ 3. Evaluate the New Limit by Direct Substitution: $$ = \frac{8^0\ln(8) - 5^0\ln(5)}{1} = \frac{1 \cdot \ln(8) - 1 \cdot \ln(5)}{1} = \ln(8) - \ln(5) $$ 4. Use Logarithm Properties: $$ = \ln\left(\frac{8}{5}\right) $$

This is the indeterminate form $\infty - \infty$. First, combine the terms into a single fraction using a common denominator. Then, apply L'Hôpital's Rule to the resulting $\frac{0}{0}$ form.

$\frac{1}{2}$

1. Combine into a Single Fraction: $$ \lim_{x \to 1^+} \left(\frac{1}{\ln(x)} - \frac{1}{x-1}\right) = \lim_{x \to 1^+} \frac{x-1 - \ln(x)}{\ln(x)(x-1)} $$ Now, as $x \to 1^+$, the numerator approaches $1-1-\ln(1) = 0$ and the denominator approaches $\ln(1)(1-1) = 0$. We have the form $\frac{0}{0}$.

2. Apply L'Hôpital's Rule: Differentiate the numerator: $1 - \frac{1}{x}$. Differentiate the denominator using the product rule: $\frac{d}{dx}(\ln(x)(x-1)) = \frac{1}{x}(x-1) + \ln(x)(1) = 1-\frac{1}{x}+\ln(x)$. $$ \lim_{x \to 1^+} \frac{1 - \frac{1}{x}}{1 - \frac{1}{x} + \ln(x)} $$ This is still a $\frac{0}{0}$ form. Applying L'Hôpital's Rule again is possible but gets complicated. Let's simplify the complex fraction first by multiplying the top and bottom by $x$: $$ \lim_{x \to 1^+} \frac{x-1}{x-1 + x\ln(x)} $$ This is still $\frac{0}{0}$. Let's apply L'Hôpital's Rule to this simpler form.

3. Apply L'Hôpital's Rule Again: Differentiate the new numerator: $1$. Differentiate the new denominator: $1 + (\ln(x) + x \cdot \frac{1}{x}) = 1 + \ln(x) + 1 = 2 + \ln(x)$. $$ \lim_{x \to 1^+} \frac{1}{2+\ln(x)} $$ 4. Evaluate the Final Limit: $$ = \frac{1}{2+\ln(1)} = \frac{1}{2+0} = \frac{1}{2} $$

This is the indeterminate form $\infty - \infty$. You can analyze this by considering the relative growth rates of the functions $e^x$ and $x$. Alternatively, factor out the dominant term, $e^x$.

$\infty$

1. Factor out the Dominant Term: The function $e^x$ grows much faster than $x$. We can factor it out to see the behavior. $$ \lim_{x \to \infty} (e^x-x) = \lim_{x \to \infty} e^x \left(1 - \frac{x}{e^x}\right) $$ 2. Analyze the Factors: We need to evaluate the limit of each part:

• $\lim_{x \to \infty} e^x = \infty$
• $\lim_{x \to \infty} \left(1 - \frac{x}{e^x}\right)$. To find $\lim_{x \to \infty} \frac{x}{e^x}$, we can use L'Hôpital's Rule on the $\frac{\infty}{\infty}$ form: $$ \lim_{x \to \infty} \frac{x}{e^x} \overset{L'H}{=} \lim_{x \to \infty} \frac{1}{e^x} = 0 $$ So, the second factor's limit is $\lim_{x \to \infty} \left(1 - \frac{x}{e^x}\right) = 1 - 0 = 1$.

3. Combine the Results: We are multiplying a term that goes to infinity by a term that goes to 1. $$ \lim_{x \to \infty} (e^x-x) \to (\infty) \cdot (1) = \infty $$