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Question 1
What indeterminate form is produced when evaluating $lim_{x \to 0} \frac{cos x - 1 + \frac{x ^{2}}{2}}{x ^{4}}$ by direct substitution?
No clue? Show me the answer
Correct answerCorrect!Incorrect
A $\frac{0}{0}$
Step-by-step walkthrough
Choose a solution method
Method #1Direct Approach
Step 1: Substitute $x = 0$ into the numerator
The numerator is $cos x - 1 + \frac{x ^{2}}{2}$ . At $x = 0$ : $cos (0) - 1 + 0 = 1 - 1 + 0 = 0$ .
Step 2: Substitute $x = 0$ into the denominator
The denominator is $x^{4}$ . At $x = 0$ : $0^{4} = 0$ .
Step 3: Identify the indeterminate form
Both numerator and denominator approach $0$ as $x \to 0$ , so the form is $\frac{0}{0}$ .
Step 4: Confirm the answer
The indeterminate form is $\frac{0}{0}$ , which is one of the two forms directly handled by L'Hôpital's Rule.
Method #2Process of Elimination
Step 1: Determine what the question is asking
We need to identify which indeterminate form arises when $x \to 0$ is substituted into the expression.
Step 2: Eliminate $\dfrac{\infty}{\infty}$
The option $\frac{\infty}{\infty}$ would require both numerator and denominator to blow up. Since $cos (0) = 1$ and $0^{4} = 0$ , there is no blow-up — this option is incorrect.
Step 3: Eliminate $0 \cdot \infty$
The form $0 \cdot \infty$ arises in products, not quotients. This expression is already a fraction, so this option does not apply.
Step 4: Eliminate $1^\infty$
The form $1^{\infty}$ arises in exponential expressions of the form $f (x)^{g (x)}$ . This expression is a simple fraction, not a power, so this option is irrelevant.
Step 5: Select $\dfrac{0}{0}$
Since both numerator ( $cos 0 - 1 + 0 = 0$ ) and denominator ( $0^{4} = 0$ ) equal zero, the correct form is $\frac{0}{0}$ .
Question 2
Using L'Hôpital's Rule, evaluate $lim_{x \to 0} \frac{e ^{3 x} - 1 - 3 x}{x ^{2}}$ .
No clue? Show me the answer
Correct answerCorrect!Incorrect
C $\frac{9}{2}$
Step-by-step walkthrough
Choose a solution method
Method #1Direct Approach
Step 1: Confirm the indeterminate form
At $x = 0$ : numerator $= e^{0} - 1 - 0 = 0$ , denominator $= 0^{2} = 0$ . Form is $\frac{0}{0}$ . ✓
Step 2: Apply L'Hôpital's Rule once
Differentiate numerator: $\frac{d}{d x} (e^{3 x} - 1 - 3 x) = 3 e^{3 x} - 3$ . Differentiate denominator: $\frac{d}{d x} (x^{2}) = 2 x$ . New limit: $lim_{x \to 0} \frac{3 e ^{3 x} - 3}{2 x}$ .
Step 3: Check if still indeterminate
At $x = 0$ : numerator $= 3 e^{0} - 3 = 0$ , denominator $= 0$ . Still $\frac{0}{0}$ , so apply L'Hôpital's Rule again.
Step 4: Apply L'Hôpital's Rule a second time
Differentiate again: numerator $\to 9 e^{3 x}$ , denominator $\to 2$ . New limit: $lim_{x \to 0} \frac{9 e ^{3 x}}{2} = \frac{9 \cdot 1}{2} = \frac{9}{2}$ .
Method #2Process of Elimination
Step 1: Recognise the structure
The expression $e^{3 x} - 1 - 3 x$ removes the first two terms of the Maclaurin series of $e^{3 x}$ . The leading surviving term is $\frac{( 3 x ) ^{2}}{2 !} = \frac{9 x ^{2}}{2}$ .
Step 2: Eliminate $0$
The answer $0$ would imply the numerator vanishes faster than $x^{2}$ , but the Maclaurin expansion shows a non-zero $x^{2}$ coefficient. This is incorrect.
Step 3: Eliminate $3$
The coefficient $3$ would arise from a linear cancellation, but after removing the $1$ and $3 x$ terms, the next term involves $\frac{9 x ^{2}}{2}$ , giving a coefficient of $\frac{9}{2}$ , not $3$ .
Step 4: Eliminate $\dfrac{3}{2}$
The value $\frac{3}{2}$ would correspond to the coefficient of $\frac{( x ) ^{2}}{2 !} = \frac{3 x ^{2}}{2}$ from $e^{x}$ , not $e^{3 x}$ . Since the exponent is $3 x$ , the coefficient is $\frac{9}{2}$ .
Step 5: Select $\dfrac{9}{2}$
The Maclaurin expansion gives $e^{3 x} - 1 - 3 x = \frac{9 x ^{2}}{2} + O (x^{3})$ , so the limit is $\frac{9 x ^{2} /2}{x ^{2}} = \frac{9}{2}$ .

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13 more questions in this topic

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Question 1
What indeterminate form is produced when evaluating $lim_{x \to 0} \frac{cos x - 1 + \frac{x ^{2}}{2}}{x ^{4}}$ by direct substitution?
No clue? Show me the answer
Correct answerCorrect!Incorrect
A $\frac{0}{0}$
Step-by-step walkthrough
Choose a solution method
Method #1Direct Approach
Step 1: Substitute $x = 0$ into the numerator
The numerator is $cos x - 1 + \frac{x ^{2}}{2}$ . At $x = 0$ : $cos (0) - 1 + 0 = 1 - 1 + 0 = 0$ .
Step 2: Substitute $x = 0$ into the denominator
The denominator is $x^{4}$ . At $x = 0$ : $0^{4} = 0$ .
Step 3: Identify the indeterminate form
Both numerator and denominator approach $0$ as $x \to 0$ , so the form is $\frac{0}{0}$ .
Step 4: Confirm the answer
The indeterminate form is $\frac{0}{0}$ , which is one of the two forms directly handled by L'Hôpital's Rule.
Method #2Process of Elimination
Step 1: Determine what the question is asking
We need to identify which indeterminate form arises when $x \to 0$ is substituted into the expression.
Step 2: Eliminate $\dfrac{\infty}{\infty}$
The option $\frac{\infty}{\infty}$ would require both numerator and denominator to blow up. Since $cos (0) = 1$ and $0^{4} = 0$ , there is no blow-up — this option is incorrect.
Step 3: Eliminate $0 \cdot \infty$
The form $0 \cdot \infty$ arises in products, not quotients. This expression is already a fraction, so this option does not apply.
Step 4: Eliminate $1^\infty$
The form $1^{\infty}$ arises in exponential expressions of the form $f (x)^{g (x)}$ . This expression is a simple fraction, not a power, so this option is irrelevant.
Step 5: Select $\dfrac{0}{0}$
Since both numerator ( $cos 0 - 1 + 0 = 0$ ) and denominator ( $0^{4} = 0$ ) equal zero, the correct form is $\frac{0}{0}$ .
Question 2
Using L'Hôpital's Rule, evaluate $lim_{x \to 0} \frac{e ^{3 x} - 1 - 3 x}{x ^{2}}$ .
No clue? Show me the answer
Correct answerCorrect!Incorrect
C $\frac{9}{2}$
Step-by-step walkthrough
Choose a solution method
Method #1Direct Approach
Step 1: Confirm the indeterminate form
At $x = 0$ : numerator $= e^{0} - 1 - 0 = 0$ , denominator $= 0^{2} = 0$ . Form is $\frac{0}{0}$ . ✓
Step 2: Apply L'Hôpital's Rule once
Differentiate numerator: $\frac{d}{d x} (e^{3 x} - 1 - 3 x) = 3 e^{3 x} - 3$ . Differentiate denominator: $\frac{d}{d x} (x^{2}) = 2 x$ . New limit: $lim_{x \to 0} \frac{3 e ^{3 x} - 3}{2 x}$ .
Step 3: Check if still indeterminate
At $x = 0$ : numerator $= 3 e^{0} - 3 = 0$ , denominator $= 0$ . Still $\frac{0}{0}$ , so apply L'Hôpital's Rule again.
Step 4: Apply L'Hôpital's Rule a second time
Differentiate again: numerator $\to 9 e^{3 x}$ , denominator $\to 2$ . New limit: $lim_{x \to 0} \frac{9 e ^{3 x}}{2} = \frac{9 \cdot 1}{2} = \frac{9}{2}$ .
Method #2Process of Elimination
Step 1: Recognise the structure
The expression $e^{3 x} - 1 - 3 x$ removes the first two terms of the Maclaurin series of $e^{3 x}$ . The leading surviving term is $\frac{( 3 x ) ^{2}}{2 !} = \frac{9 x ^{2}}{2}$ .
Step 2: Eliminate $0$
The answer $0$ would imply the numerator vanishes faster than $x^{2}$ , but the Maclaurin expansion shows a non-zero $x^{2}$ coefficient. This is incorrect.
Step 3: Eliminate $3$
The coefficient $3$ would arise from a linear cancellation, but after removing the $1$ and $3 x$ terms, the next term involves $\frac{9 x ^{2}}{2}$ , giving a coefficient of $\frac{9}{2}$ , not $3$ .
Step 4: Eliminate $\dfrac{3}{2}$
The value $\frac{3}{2}$ would correspond to the coefficient of $\frac{( x ) ^{2}}{2 !} = \frac{3 x ^{2}}{2}$ from $e^{x}$ , not $e^{3 x}$ . Since the exponent is $3 x$ , the coefficient is $\frac{9}{2}$ .
Step 5: Select $\dfrac{9}{2}$
The Maclaurin expansion gives $e^{3 x} - 1 - 3 x = \frac{9 x ^{2}}{2} + O (x^{3})$ , so the limit is $\frac{9 x ^{2} /2}{x ^{2}} = \frac{9}{2}$ .

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Step-by-step walkthrough

Step 1: Substitute $x = 0$ into the numerator

Step 2: Substitute $x = 0$ into the denominator

Step 3: Identify the indeterminate form

Step 4: Confirm the answer

Step 1: Determine what the question is asking

Step 2: Eliminate $\dfrac{\infty}{\infty}$

Step 3: Eliminate $0 \cdot \infty$

Step 4: Eliminate $1^\infty$

Step 5: Select $\dfrac{0}{0}$

Step-by-step walkthrough

Step 1: Confirm the indeterminate form

Step 2: Apply L'Hôpital's Rule once

Step 3: Check if still indeterminate

Step 4: Apply L'Hôpital's Rule a second time

Step 1: Recognise the structure

Step 2: Eliminate $0$

Step 3: Eliminate $3$

Step 4: Eliminate $\dfrac{3}{2}$

Step 5: Select $\dfrac{9}{2}$

13 more questions in this topic

Step-by-step walkthrough

Step 1: Substitute $x = 0$ into the numerator

Step 2: Substitute $x = 0$ into the denominator

Step 3: Identify the indeterminate form

Step 4: Confirm the answer

Step 1: Determine what the question is asking

Step 2: Eliminate $\dfrac{\infty}{\infty}$

Step 3: Eliminate $0 \cdot \infty$

Step 4: Eliminate $1^\infty$

Step 5: Select $\dfrac{0}{0}$

Step-by-step walkthrough

Step 1: Confirm the indeterminate form

Step 2: Apply L'Hôpital's Rule once

Step 3: Check if still indeterminate

Step 4: Apply L'Hôpital's Rule a second time

Step 1: Recognise the structure

Step 2: Eliminate $0$

Step 3: Eliminate $3$

Step 4: Eliminate $\dfrac{3}{2}$

Step 5: Select $\dfrac{9}{2}$

13 more questions in this topic