This problem applies L'Hôpital's Rule to the limit as x approaches infinity of ln(x)/√x. As x goes to infinity, it's ∞/∞, allowing the rule. Derivatives are (1/x)/(1/(2√x)) = 2/√x, which approaches 0. Logarithmic growth is slower than any positive power root. A tempting distractor is ∞, mistakenly thinking ln x grows faster than √x. To recognize indeterminate forms at infinity, assess if both numerator and denominator tend to infinity or zero in a way that creates ambiguity.

This problem applies L'Hôpital's Rule to the limit as x approaches 0 of $\frac{\sqrt{1+9x} - 1}{x}$. It results in $\frac{1-1}{0} = \frac{0}{0}$, justifying the rule. Derivative is $\frac{9}{2\sqrt{1+9x}} / 1$, evaluating to $\frac{9}{2}$ at x=0. Rationalizing by multiplying conjugate also yields $\frac{9}{2}$. A tempting distractor is $9$, from forgetting the 1/2 in the square root derivative. To identify $0/0$ forms, plug in the limit value and ensure it's indeterminate before differentiating.

This problem uses L'Hôpital's Rule to find the limit as x approaches 0 of (x - sin $x)/x^3$. Substituting gives 0/0, an indeterminate form for the rule. First derivatives (1 - cos $x)/(3x^2$) are 0/0; second (sin x)/(6x) are 0/0; third (cos x)/6 = 1/6 at x=0. This aligns with sin x ≈ x - $x^3$/6 from Taylor series. A tempting distractor is -1/6, possibly from sign error in series. Recognize persistent indeterminate forms by applying the rule multiple times until resolved.

L'Hôpital's Rule evaluates the limit of $ \frac{\tan x - x}{x^3} $ as x approaches 0. It's $ \frac{0}{0} $, first derivatives $ \sec^2 x - \frac{1}{3x^2} $, still $ \frac{0}{0} $. Second: $ \frac{2 \sec^2 x \tan x}{6x} $, $ \frac{0}{0} $. Third: complex but evaluates to $ \frac{1}{3} $. Repeated indeterminates justify the process. The distractor $ 0 $ might come from premature stopping. Spot indeterminate forms by checking limits of numerator and denominator independently.

This limit problem requires L'Hôpital's Rule to evaluate an indeterminate form. As x approaches 0, sin(2x) - 2 sin(x) approaches 0 - 0 = 0 and $x^3$ approaches 0, yielding 0/0. Derivatives: 2 cos(2x) - 2 cos(x) over $3x^2$, still 0/0 (21 - 21=0); again: -4 sin(2x) + 2 sin(x) over 6x, still 0/0; again: -8 cos(2x) + 2 cos(x) over 6 = (-81 + 21)/6 = -6/6 = -1. Repeated application is needed due to persistent indeterminacy. A tempting distractor is 0, from direct substitution without accounting for higher derivatives. To recognize indeterminate forms like 0/0, always plug in the limit value to check if both numerator and denominator approach 0 or infinity.

L'Hôpital's Rule is essential for the limit of (sin x - $x)/x^3$ as x approaches 0. Substitution yields 0/0, leading to derivatives cos x - 1 over $3x^2$, still 0/0. A second application gives -sin x / (6x), again 0/0, and a third gives -cos x / 6 = -1/6. The persistent indeterminate form justifies repeated use of the rule. The distractor 0 might arise from stopping after the first differentiation. To spot indeterminate forms, evaluate step-by-step after each rule application.

L'Hôpital's Rule helps evaluate the limit of tan(2x)/x as x approaches 0. Substitution gives 0/0, so we differentiate to 2 $sec^2$(2x) over 1, which is 2 at x=0. The indeterminate form justifies applying the rule to find this finite limit. No additional applications are required here. The distractor 1 might tempt if one forgets the factor of 2 from the chain rule. A strategy for spotting indeterminate forms is to compute numerator and denominator separately at the limit value.

L'Hôpital's Rule is the key skill for computing the limit of $(e^{4x} - 1)/x$ as x approaches 0. Substituting x=0 produces $0/0$, an indeterminate form that allows us to apply the rule by differentiating numerator and denominator. The derivative of the numerator is $4e^{4x}$ and of the denominator is 1, yielding $4e^{0} = 4$. This direct application resolves the limit without further steps. The distractor $e^4$ might arise from incorrectly exponentiating instead of differentiating properly. Always check for $0/0$ or $\infty/\infty$ by plugging in the limit value to confirm if L'Hôpital's Rule applies.

This limit problem requires L'Hôpital's Rule to evaluate an indeterminate form. As x approaches 0, both the numerator sin(x) and the denominator √(1+x) - 1 approach 0, creating a 0/0 indeterminate form that justifies differentiating the numerator and denominator. The derivative of the numerator is cos(x), and the derivative of the denominator is 1/(2√(1+x)), so the limit becomes $lim_{x→0}$ cos(x) / (1/(2√(1+x))) = 1 / (1/2) = 2. Applying L'Hôpital's Rule resolves the indeterminacy because the derivatives exist and the limit of the ratio exists. A tempting distractor might be 1/2, which could arise from mistakenly rationalizing the denominator without applying L'Hôpital correctly. To recognize indeterminate forms like 0/0, always plug in the limit value to check if both numerator and denominator approach 0 or infinity.

This limit problem requires L'Hôpital's Rule to evaluate an indeterminate form. As x approaches 0, ln(1+x) approaches 0 and √(1+x) - 1 approaches 0, resulting in a 0/0 form that allows us to apply the rule. Differentiating gives 1/(1+x) in the numerator and 1/(2√(1+x)) in the denominator, so the limit is $lim_{x→0}$ [1/(1+x)] / [1/(2√(1+x))] = 1 / (1/2) = 2. The rule is justified here because the original limit is indeterminate, but the derivatives lead to a determinate value. One tempting distractor is 1/2, possibly from inverting the derivative ratio incorrectly. To recognize indeterminate forms like 0/0, always plug in the limit value to check if both numerator and denominator approach 0 or infinity.