$0/0$ 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.

$0/0$; 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.

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.

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 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.

This limit has the indeterminate form 0/0 since both tan(x) - x and x³ approach 0 as x → 0. Applying L'Hôpital's Rule once gives (sec²(x) - 1)/(3x²), which can be rewritten as tan²(x)/(3x²) using the identity sec²(x) - 1 = tan²(x). This is still 0/0, so applying L'Hôpital's Rule again yields 2tan(x)sec²(x)/(6x). One more application gives 2sec²(x)(sec²(x) + 2tan²(x)sec²(x))/6, which evaluates to 2(1)(1 + 0)/6 = 1/3 as x → 0. A common error is not recognizing that sec²(x) - 1 = tan²(x), which simplifies the calculation. For functions like tan(x) - x near x = 0, expect the leading term in the Taylor expansion to be x³/3.

Indeterminate 0/0; L'Hôpital gives (2x $e^{x²}$)/(2x) = $e^{x²}$ → 1. Equivalent to $(e^v$ -1)/v → 1, v=x². Tempting wrong: assuming 0. Another error: wrong derivative. Strategy: recognize standard limits in disguised forms.

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.

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.

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.