L'Hospital's Rule - AP Calculus AB

A growth comparison uses $\lim_{x\to\infty}\frac{\ln x}{\sqrt{x}}$ . What is the value of the limit?
Answer: $0$ This limit has the indeterminate form ∞/∞ as both ln(x) and √x approach infinity as x → ∞. Applying L'Hôpital's Rule gives (1/x)/(1/(2√x)) = 2√x/x = 2/√x. As x → ∞, this expression approaches 0 since the denominator grows without bound. A common mistake is to incorrectly differentiate √x as 1/√x instead of 1/(2√x), which would lead to the wrong answer. The key insight is that logarithmic functions grow much more slowly than any positive power of x, so ln(x)/ $x^p$ → 0 as x → ∞ for any p > 0.
Compute $\displaystyle \lim_{x\to 0}\frac{e^{x}-1-\sin x}{x^2}$ .
Answer: $\dfrac{1}{2}$ This limit is 0/0 indeterminate, so L'Hôpital's Rule applies. Differentiating gives ( $e^x$ - cos x)/(2x), still 0/0; a second differentiation yields ( $e^x$ + sin x)/2, evaluating to (1 + 0)/2 = 1/2. Taylor series confirms: $e^x$ ≈ 1 + x + x²/2, sin x ≈ x, so numerator ≈ x²/2, divided by x² gives 1/2. A common mistake is to differentiate incorrectly, perhaps forgetting the chain rule for $e^x$ . Another tempting error is assuming the limit is 0 by direct substitution without accounting for rates. A transferable strategy is to apply L'Hôpital's Rule until the form is determinate and cross-verify with series expansions for conceptual understanding.
Find $\displaystyle \lim_{x\to 0}\frac{1-\cos(9x)}{x^2}$
Answer: $\dfrac{81}{2}$ Lim $\frac{1 - \cos(9x)}{x^2}$ as $x \to 0$ is 0/0, apply L'Hôpital twice. Derivatives lead to $\frac{81}{2}$ via scaling the standard $\frac{1}{2}$ limit by $9^2$ . A wrong move is forgetting the square, getting 9. The conceptual step uses $\frac{1 - \cos(kx)}{x^2} = \frac{k^2}{2}$ . This generalizes trigonometric limits. A transferable strategy is leveraging known limits with substitutions for efficiency.
Evaluate $\displaystyle \lim_{x\to 0}\frac{\ln(1+x)-x}{x^2}$ .
Answer: $-\dfrac{1}{2}$ The limit (ln(1+x) - x)/ $x^2$ as x→0 is 0/0, use L'Hôpital. Derivatives: 1/(1+x) - 1 / 2x, 0/0; second: -1/ $(1+x)^2$ / 2 = -1/2. One might incorrectly assume it's 0 from direct substitution. The step involves Taylor: ln(1+x) ≈ x - $x^2$ /2, yielding -1/2. This captures the quadratic term. A transferable strategy is applying L'Hôpital for logs and confirming with series for signs.
A decay model requires $\lim_{x\to0}\frac{\ln(1+7x)-7x}{x^2}$ . What is the limit?
Answer: $-\frac{49}{2}$ The expression ln(1 + 7x) - 7x approaches 0 as x approaches 0, creating a 0/0 indeterminate form. Applying L'Hôpital's Rule once gives (7/(1 + 7x) - 7)/(2x) = -49x/((1 + 7x)(2x)), which is still 0/0. Applying it again yields -49/(2(1 + 7x)) - 343x/(2(1 + 7x)²), which equals -49/2 when x = 0. A common error is forgetting that ln(1 + u) ≈ u - u²/2 for small u, which explains why the second-order term dominates. The strategy is to remember that when logarithms appear with their linear approximations subtracted, the limit often involves the coefficient squared divided by 2.
Compute $\displaystyle \lim_{x\to 0}\frac{\ln(1+2x)-2x}{x^2}$ .
Answer: $-2$ Lim (ln(1+2x) - 2x)/ $x^2$ is 0/0, L'Hôpital twice gives -2. A mistake is positive sign assumption. Taylor: - ( $4x^2$ )/2 = - $2x^2$ . The step captures negative quadratic. It evaluates correctly to -2. A transferable strategy is mindful of series signs in logs.
Compute $\displaystyle \lim_{x\to 0}\frac{e^{\sin x}-1-\sin x}{x^2}$ .
Answer: $\dfrac{1}{2}$ 0/0; twice L'Hôpital gives 1/2. Taylor: $e^{sin x}$ -1 - sin x ≈ (sin x)²/2 / x² →1/2. Wrong: approximating e as 1+sin x only. Error: derivative mishap. Strategy: expand exponentials with inner series.
To compare growth rates, evaluate $\lim_{x\to\infty}\frac{\ln x}{\sqrt{x}}$ using L'Hospital's Rule.
Answer: $0$ As x approaches infinity, both $\ln(x)$ and $\sqrt{x}$ approach infinity, giving the indeterminate form $\infty / \infty$ . Applying L'Hôpital's Rule, we differentiate to get $\frac{1/x}{1/(2\sqrt{x})} = \frac{2\sqrt{x}}{x} = \frac{2}{\sqrt{x}}$ . As x approaches infinity, $\frac{2}{\sqrt{x}}$ approaches 0, showing that logarithmic growth is slower than square root growth. A common mistake is to think that since both functions grow without bound, the limit doesn't exist or equals 1. The key insight is that L'Hôpital's Rule helps compare growth rates: when the limit is 0, the numerator grows more slowly than the denominator.
Compute $\displaystyle \lim_{x\to 0}\frac{\ln(1+x^2)}{x^2}$ .
Answer: $1$ As x approaches 0, it's 0/0, so L'Hôpital's Rule: derivative (2x/(1+x²))/(2x) = 1/(1+x²) → 1. Known limit ln(1+u)/u → 1 with u = x². A common mistake is direct substitution giving 0/0 without applying the rule. Another tempting error is confusing with other logs. A transferable strategy is substituting variables for composition limits.
Compute $\displaystyle \lim_{x\to 0}\frac{\ln(1+3x)-3x}{x^2}$ .
Answer: $-\dfrac{9}{2}$ Evaluating lim (ln(1+3x) - 3x)/ $x^2$ as x→0 gives 0/0, requiring L'Hôpital's Rule. Derivatives are 3/(1+3x) - 3 over 2x, still 0/0; second derivatives yield -9/ $(1+3x)^2$ over 2, which is -9/2 at x=0. One might wrongly plug in after the first step, thinking it's resolved. The key is recognizing the need for two applications due to the quadratic behavior. Taylor series for ln(1+3x) confirms the -9/2 limit. A transferable strategy is to use L'Hôpital's Rule for logarithmic indeterminates, applying it multiple times and cross-checking with series expansions.
Find $\displaystyle \lim_{x\to 0}\frac{\sin x - x\,e^{x}}{x^2}$ .
Answer: $-1$ An indeterminate 0/0 form arises, so apply L'Hôpital's Rule. First derivative: (cos x - $e^x$ - x $e^x$ )/ (2x), still 0/0; second: (-sin x - $e^x$ - $e^x$ - x $e^x$ )/2 = (-sin x - $2e^x$ - x $e^x$ )/2, evaluating to (-0 - 2 - 0)/2 = -1. Taylor series: sin x ≈ x - x³/6, x $e^x$ ≈ x + x² + (x³)/2, so numerator ≈ -x² - (2/3)x³, divided by x² gives -1 + higher terms → -1. A common mistake is incorrect expansion of $e^x$ terms. Another tempting error is direct substitution leading to 0. A transferable strategy is combining L'Hôpital's Rule with series for products like x $e^x$ in numerators.
Evaluate $\displaystyle \lim_{x\to 0}\frac{1-\cos(x^2)}{x^4}$ .
Answer: $\dfrac{1}{2}$ $0/0$ ; twice L'Hôpital: derivative $\sin(x^2) \cdot 2x / 4x^3$ , adjust; better identity $1 - \cos u = \frac{u^2}{2}$ approx, u= $x^2$ , gives $\frac{1}{2}$ . Mistake: using $1 - \cos x$ . Error: insufficient applications. Strategy: use trig identities before rule.
A log-growth index is $\lim_{x\to\infty}\frac{\ln x}{x^{1/3}}$ . What is the limit?
Answer: $0$ This limit has the indeterminate form ∞/∞ as both ln(x) and $x^{1/3}$ approach infinity as x → ∞. Applying L'Hôpital's Rule gives (1/x)/((1/3) $x^{-2/3}$ ) = $3x^{-2/3}$ /x = $3x^{-2/3-1}$ = $3x^{-5/3}$ = 3/ $x^{5/3}$ . As x → ∞, this expression approaches 0 since the denominator grows without bound. A common mistake is incorrectly differentiating $x^{1/3}$ , forgetting to apply the power rule properly. The general principle is that ln(x) grows more slowly than any positive power of x, no matter how small that power is.
Evaluate $\displaystyle \lim_{x\to 0}\frac{1-\cos(2x)}{x^2}$ .
Answer: $2$ The limit $\frac{1 - \cos(2x)}{x^2}$ is $0/0$ , L'Hôpital twice or standard form gives $2$ . Temptingly, ignoring coefficient yields $\frac{1}{2}$ . The step scales by $4/2 = 2$ . This uses trig identity. Verification is $2$ . A transferable strategy is coefficient adjustment in limits.
Find $\displaystyle \lim_{x\to 0}\frac{\sin(x^2)}{x^2}$ .
Answer: $1$ $0/0$ form; L'Hôpital: $\frac{2x \cos(x^2)}{2x} = \cos(x^2) \to 1$ . Or $\frac{\sin(u)}{u} \to 1$ , $u = x^2$ , adjusted. Mistake: thinking it's $\sin x / x$ . Another: substitution error. Strategy: variable substitution for inner functions.
Evaluate $\displaystyle \lim_{x\to 0}\frac{e^{x}-1-x}{x^2}$ .
Answer: $\dfrac{1}{2}$ Lim ( $e^x$ - 1 - x)/ $x^2$ as x→0 is 0/0, L'Hôpital twice yields 1/2. A temptation is confusing with first-order approximation. The correct step uses $e^x$ Taylor: $x^2$ /2 term. This highlights second-order expansion. Verification matches 1/2. A transferable strategy is systematic Taylor use for exponentials.
As $x\to 0$ , a response factor is $\frac{e^x-1-x}{x^2}$ . What is the limit?
Answer: $\dfrac{1}{2}$ This limit presents the indeterminate form 0/0 as both $e^x$ - 1 - x and x² approach 0 when x → 0. Applying L'Hôpital's Rule once gives ( $e^x$ - 1)/(2x), which is still 0/0. Applying it again yields $e^x$ /2, which evaluates to 1/2 as x → 0. Students often forget to differentiate the -x term in the first application, leading to errors. This result can also be verified using the Taylor expansion $e^x$ = 1 + x + x²/2 + ..., which shows that $e^x$ - 1 - x = x²/2 + higher order terms, confirming the limit is 1/2.
For $x\to 0$ , the efficiency model is $\frac{e^{2x}-1-2x}{x^2}$ . What is the limit?
Answer: $2$ This limit presents the indeterminate form 0/0 as both $e^{2x}$ - 1 - 2x and x² approach 0 when x → 0. Applying L'Hôpital's Rule once gives ( $2e^{2x}$ - 2)/(2x), which simplifies to ( $e^{2x}$ - 1)/x, still 0/0. Applying L'Hôpital's Rule again yields $2e^{2x}$ /1 = $2e^{2x}$ , which evaluates to 2 as x → 0. Students often make the error of forgetting to differentiate the constant term -2x in the numerator, leading to incorrect results. For exponential expressions of the form $e^{ax}$ - 1 - ax, the limit as x → 0 of the expression divided by x² is always a²/2.
Find $\displaystyle \lim_{x\to 0}\frac{\ln(1+x)-\sin x}{x^2}$ .
Answer: $-\dfrac{1}{2}$ The limit $\frac{\ln(1+x) - \sin x}{x^2}$ is $0/0$ , L'Hôpital twice gives $-1/2$ . Wrongly adding terms positively. Taylor: $-x^2/2$ . The step notes opposing series. It evaluates to $-1/2$ . A transferable strategy is combining different function series.
For a long-term prediction, find $\lim_{x\to\infty}\frac{2x^2+5}{e^x}$ using L'Hospital's Rule.
Answer: $0$ This limit has the indeterminate form ∞/∞ as both the numerator $2x^2 + 5$ and denominator $e^x$ approach infinity. Applying L'Hôpital's Rule once gives $\frac{4x}{e^x}$ , still ∞/∞. Applying it again yields $4/e^x$ , which approaches 0 as x approaches infinity. A third application would give $0/e^x$ = 0, confirming the result. The key insight is that exponential functions grow faster than any polynomial, so $e^x$ eventually dominates any polynomial in the numerator. The transferable strategy is that when comparing polynomial and exponential growth, the exponential always wins, making such limits equal to 0.

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