Trap 1: Degree confusion in x -> -infinity lim(x->-infinity) sqrt($x^2$ + x) + x. Students write sqrt($x^2$) = x. Wrong! For x < 0, sqrt($x^2$) = -x. Correct: sqrt($x^2$ + x) = |x| * sqrt(1 + 1/x) = -x * sqrt(1 + 1/x) when x < 0. Then expression = -x * sqrt(1 + 1/x) + x = x(-sqrt(1+1/x) + 1). As x -> -infinity, this -> -infinity * 0, need careful expansion. Using conjugate: answer = -1/2.

Trap 2: sin$\frac{x}{x}$ when x is in degrees lim(x->0) sin$\frac{x degrees}{x}$ = sin$\frac{pi*x/180}{(x)}$ -> pi/180, NOT 1.

Trap 3: [sin x / x] as x -> 0 sin x / x -> 1 from below (for small x > 0, sin x < x). So [sin x / x] = 0 near 0, not 1.

Trap 4: Misapplying L'Hopital lim(x->0) $\frac{sin x}{x}$ is 1 by standard limits. Using L'Hopital: cos$\frac{0}{1}$ = 1. Both work, but don't use L'Hopital on lim(x->0) (x * sin$\frac{1}{x}$) — this is NOT 0/0 form, it's 0 * bounded.

Trap 5: Canceling before checking lim(x->1) $\frac{x^2-1}{(x-1)}$ = lim (x+1) = 2. Correct. But only valid because we cancel (x-1) and then substitute. Don't substitute x=1 before canceling.