Common limits, with the method

A solver returns the number. Each page here also names WHICH technique the limit wants and why the obvious first move fails, because choosing the method is the part students actually get stuck on.

• $\lim_{x \to 0} \frac{\sin x}{x} = 1$the special trigonometric limit
• $\lim_{x \to 0} \frac{1-\cos x}{x} = 0$the special trigonometric limit
• $\lim_{x \to 0} \frac{1-\cos x}{x^2} = \frac{1}{2}$L'Hopital's rule applied twice
• $\lim_{x \to 0} \frac{\tan x}{x} = 1$rewriting as sine over cosine
• $\lim_{x \to 0} \frac{\sin 2x}{x} = 2$matching the special trigonometric limit
• $\lim_{x \to 0} \frac{\sin 3x}{\sin 5x} = \frac{3}{5}$the special trigonometric limit applied twice
• $\lim_{x \to 0} \frac{\tan x}{\sin x} = 1$algebraic simplification
• $\lim_{x \to 0} \frac{e^x-1}{x} = 1$the limit definition of the derivative
• $\lim_{x \to 1} \frac{x^2-1}{x-1} = 2$factoring and cancelling
• $\lim_{x \to 2} \frac{x^3-8}{x-2} = 12$factoring the difference of cubes
• $\lim_{x \to 2} \frac{x^2-4}{x^2-x-2} = \frac{4}{3}$factoring both polynomials
• $\lim_{x \to 4} \frac{\sqrt{x}-2}{x-4} = \frac{1}{4}$multiplying by the conjugate
• $\lim_{x \to \infty} \left(1+\frac{1}{x}\right)^x = e$the definition of e
• $\lim_{x \to 0^+} (1+x)^{1/x} = e$the definition of e
• $\lim_{x \to 0^+} x^x = 1$logarithms then L'Hopital
• $\lim_{x \to 0^+} x\ln x = 0$rewriting as a quotient then L'Hopital
• $\lim_{x \to 0^+} \ln x = -\infty$end behaviour of the logarithm
• $\lim_{x \to 0} \frac{1}{x^2} = \infty$unbounded behaviour at a vertical asymptote
• $\lim_{x \to \infty} \frac{1}{x} = 0$end behaviour
• $\lim_{x \to \infty} \frac{2x^2+3}{x^2-1} = 2$comparing leading degrees
• $\lim_{x \to \infty} \frac{3x+1}{x^2+2} = 0$comparing leading degrees
• $\lim_{x \to \infty} \frac{x^2}{x+1} = \infty$comparing leading degrees
• $\lim_{x \to -\infty} e^x = 0$end behaviour of the exponential
• $\lim_{x \to \infty} \frac{\ln x}{x} = 0$L'Hopital's rule
• $\lim_{x \to \infty} \frac{\sin x}{x} = 0$the squeeze theorem
• $\lim_{x \to \infty} \arctan x = \frac{\pi}{2}$the range of the arctangent
• $\lim_{x \to \infty} \frac{x}{\sqrt{x^2+1}} = 1$dividing by the dominant term
• $\lim_{x \to \infty} \sqrt{x^2+x}-x = \frac{1}{2}$multiplying by the conjugate