This problem requires the use of L'Hôpital's Rule to evaluate the limit. As t approaches 0, both sin(5t) and t approach 0, creating the indeterminate form 0/0, which allows us to apply L'Hôpital's Rule by differentiating the numerator and denominator. The derivative of the numerator is 5 cos(5t), and the derivative of the denominator is 1, so the limit becomes 5 cos(5t) as t approaches 0. This evaluates to 5 cos(0) = 5 * 1 = 5. A tempting distractor is 1/5, which might result from mistakenly inverting the coefficient in the standard sin(u)/u limit without proper adjustment. To recognize indeterminate forms transferably, always substitute the limiting value into the numerator and denominator separately to check for 0/0 or ∞/∞.

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 involves applying L'Hôpital's Rule to evaluate the limit as $t \to 0$ of $\dfrac{e^{5t} - 1}{t}$. As $t$ approaches 0, the numerator $e^{5t} - 1$ approaches 0 and the denominator $t$ approaches 0, creating the indeterminate form $\frac{0}{0}$ that justifies using L'Hôpital's Rule. Differentiating the numerator gives $5e^{5t}$ and the denominator gives 1, so the limit simplifies to $5e^{5t}$ evaluated at $t=0$, which is 5. This result aligns with the derivative of $e^{5t}$ at $t=0$, providing a conceptual check. A tempting distractor is $e^5$, which might arise from mistakenly evaluating the numerator at $t=5$ instead of applying the rule properly. To recognize indeterminate forms like $\frac{0}{0}$ or $\frac{\infty}{\infty}$ in limits, substitute the limiting value and verify if the expression is undefined.

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 applies L'Hôpital's Rule to evaluate the limit as x approaches 0 of $\frac{1 - \cos(6x)}{x^2}$. Substituting x=0 gives $\frac{1-1}{0} = \frac{0}{0}$, an indeterminate form that permits the rule. First derivatives are $\frac{6 \sin(6x)}{2x}$, still $\frac{0}{0}$, so apply again to get $\frac{36 \cos(6x)}{2} = 18 \cos(0) = 18$. Multiple applications are needed due to the persistent indeterminate form. A tempting distractor is 36, which might come from forgetting to divide by 2 after the second differentiation. Recognize indeterminate forms by substituting the limit value and repeatedly checking after each differentiation if needed.

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.

This problem requires L'Hôpital's Rule to find the limit as x approaches 0 of $\sin(7x)/x$. Substituting x=0 yields $\sin(0)/0 = 0/0$, an indeterminate form that allows L'Hôpital's Rule to be applied. The derivatives are $7 \cos(7x)$ for the numerator and $1$ for the denominator, so the limit is $7 \cos(0) = 7$. This is equivalent to 7 times the standard limit of $\sin(u)/u$ as u approaches 0, where $u=7x$. A tempting distractor is 1, which ignores the coefficient 7 and treats it as the basic sine limit. Always check for indeterminate forms by plugging in the limit value and seeing if it results in $0/0$ or $\infty/\infty$ before proceeding with differentiation.

This problem involves L'Hôpital's Rule for the limit as x approaches infinity of $\frac{e^{2x}}{x}$. It presents as $\infty/\infty$, justifying application. Derivative is $\frac{2e^{2x}}{1}$, still $\infty/\infty$, and repeated applications show exponential dominance, yielding $\infty$. Exponentials grow faster than polynomials. A tempting distractor is 0, from incorrectly assuming the denominator overtakes. Spot indeterminate $\infty/\infty$ forms by evaluating asymptotic behavior and confirming unbounded growth in both parts.

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.