$0/0$ 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.

0/0; L'Hôpital: (2x sec²(x²))/(2x) = sec²(x²) →1. Similar to tan u /u →1. Wrong: confusing with tan x /x. Error: incorrect chain rule. Strategy: apply known limits to composites.

This limit has the indeterminate form $0/0$ since $e^{4x} - 1 - 4x$ approaches 0 as x approaches 0 (the linear approximation of $e^{4x}$ near 0 is $1 + 4x$). Applying L'Hôpital's Rule once gives $(4e^{4x} - 4)/(2x)$, which is still $0/0$. Applying it again yields $16e^{4x}/2 = 8e^{4x}$, which equals $8$ when x = 0. A tempting error is to think that $e^{4x} - 1 \approx 4x$ near zero, making the numerator approximately 0, but this ignores the quadratic term in the Taylor expansion. The strategy is to recognize that when exponential functions appear with their linear approximations subtracted, you often need L'Hôpital's Rule twice to resolve the indeterminacy.

As $x$ approaches infinity, both $x$ and $\ln(x^2 + 1)$ approach infinity, creating an $\infty / \infty$ indeterminate form. Applying L'Hôpital's Rule gives $\frac{1}{\frac{2x}{x^2 + 1}} = \frac{x^2 + 1}{2x}$. As $x$ approaches infinity, this simplifies to $\frac{x^2}{2x} = \frac{x}{2}$, which approaches infinity. The limit is therefore infinity, not a finite value. A common mistake is to think that since ln grows slowly, the limit might be finite, but x grows faster than $\ln(x^2 + 1) \approx 2\ln(x)$. The strategy is to simplify after each application of L'Hôpital's Rule and check whether the result still approaches infinity.

As t approaches 0, both the numerator sin(5t) - 5t and denominator t³ approach 0, creating a 0/0 indeterminate form that requires L'Hôpital's Rule. Applying L'Hôpital's Rule once gives (5cos(5t) - 5)/(3t²), which is still 0/0 at t = 0. Applying it again yields (-25sin(5t))/(6t), still 0/0. A third application gives (-125cos(5t))/6, which evaluates to -125/6 when t = 0. A common mistake is to stop after one or two applications when the form is still indeterminate. The key strategy is to continue applying L'Hôpital's Rule until you get a determinate form, checking the limit after each differentiation.

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.

The limit as t approaches 0 of $(e^{5t}$ - 1 - $5t)/t^2$ results in the indeterminate form 0/0, necessitating L'Hôpital's Rule. Differentiating the numerator gives $5e^{5t}$ - 5 and the denominator 2t, which still yields 0/0. Applying L'Hôpital's Rule again produces $25e^{5t}$ in the numerator and 2 in the denominator, evaluating to 25/2 at t=0. A tempting wrong approach is to mistakenly conclude the limit is 0 after the first differentiation without checking the form. Instead, recognize that multiple applications may be needed for higher-order terms. A transferable strategy is to apply L'Hôpital's Rule repeatedly until the indeterminate form is resolved, ensuring each step is verified.

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.

The limit (sin x - x cos $x)/x^3$ is 0/0, but one L'Hôpital gives (x sin x)/ $(3x^2$) = sin x /(3x) → 1/3. Wrongly applying more times complicates unnecessarily. The step simplifies after first differentiation to a known limit. Taylor confirms $x^3$/3. This shows not always needing max applications. A transferable strategy is checking if form resolves early.

The limit $(e^{3x}$ - 1 - $3x)/x^2$ is 0/0, L'Hôpital twice gives 9/2. Mistakenly halving incorrectly yields 9. Taylor: $(9x^2$)/2 term. The step identifies second-order. It matches 9/2. A transferable strategy is exponential series for quick limits.