Evaluating Limits — Factor, Rationalise, Standard Forms

$\lim_{x\to a}f(x)=L$ means $f(x)$ gets arbitrarily close to $L$ as $x\to a$. It exists iff LHL $=$ RHL. Algebra of limits: limits distribute over sums, products, and quotients (when denominators are non-zero). For $n\to\infty$ ratios, the dominant term decides (e.g. $\dfrac{a^n+b^n}{a^{n}}\to$ the larger base's contribution).

• The dominant-term rule, written out: $r^n\to0$ for $|r|<1$. So for $a>b>0$, divide by $a^n$: $\dfrac{a^n+b^n}{a^n-b^n}=\dfrac{1+(b/a)^n}{1-(b/a)^n}\to\dfrac{1+0}{1-0}=1$; and $\dfrac{2^n+3^n}{3^n}\to1$, $\dfrac{2^n}{3^n}\to0$.
• The algebra of limits does NOT run backwards: $\lim(fg)$ can exist while neither $\lim f$ nor $\lim g$ does ($f=\tfrac1x$, $g=x$ at 0), and $\lim(f+g)$ can exist with both summands divergent ($f=\tfrac1x$, $g=-\tfrac1x$). "If $\lim fg$ exists then $\lim f$ exists" is the planted false statement.
• A limit that IS a derivative: $\lim_{h\to0}\dfrac{f(a+h)-f(a)}{h}=f'(a)$ and $\lim_{x\to a}\dfrac{f(x)-f(a)}{x-a}=f'(a)$. Recognise the shape and differentiate instead of manipulating: $\lim_{x\to a}\dfrac{x^n-a^n}{x-a}=na^{n-1}$, $\lim_{x\to0}\dfrac{\cos x-1}{x}=(\cos)'(0)=0$, $\lim_{x\to 2}\dfrac{e^{x}-e^{2}}{x-2}=e^{2}$.
• The vanishing-numerator rule: if $\lim\dfrac{f}{g}$ is FINITE and $g\to0$, then $f\to0$ too — so "$\lim_{x\to1}\dfrac{x^2+ax+b}{x-1}=5$" forces $1+a+b=0$ first, and then the value pins the second unknown ($2+a=5$).

• Factor/cancel: use $x^n-a^n=(x-a)(x^{n-1}+\cdots+a^{n-1})$; the standard result $\lim_{x\to a}\dfrac{x^n-a^n}{x-a}=n\,a^{n-1}$.
• Rationalise: multiply numerator and denominator by the conjugate of the surd to turn $\sqrt{A}-\sqrt{B}$ into $A-B$, then cancel.
• A quadratic from its roots: if the roots are $m,n$ then $x^2-(m+n)x+mn=(x-m)(x-n)$ — the factorisation a 0/0 limit at $x=m$ needs, without solving anything.
• A hidden perfect square: $a^x+a^{-x}-2=\big(a^{x/2}-a^{-x/2}\big)^2\ge0$, so the numerator tends to 0 from above on both sides — which is what separates the two one-sided limits of $\dfrac{\sqrt{a^x+a^{-x}-2}}{x}$ ($\pm\ln a$). Likewise $1-\cos x=2\sin^2\tfrac x2\ge0$.
• The $\infty-\infty$ form ($x\to\infty$): combine into ONE fraction, or rationalise at infinity — $\sqrt{x^2+x}-x=\dfrac{x}{\sqrt{x^2+x}+x}=\dfrac{1}{\sqrt{1+1/x}+1}\to\tfrac12$. Never subtract the two infinities term by term; the answer depends on how fast each grows.

Each holds as $x\to 0$ (angles in radians). For a scaled argument, the factor scales too — e.g. $\lim_{x\to0}\dfrac{\sin ax}{x}=a$.

If $\lim\dfrac{f}{g}$ is $\tfrac00$ or $\tfrac{\infty}{\infty}$ and $f,g$ are differentiable, then $\lim\dfrac{f}{g}=\lim\dfrac{f'}{g'}$ (provided the latter exists). Only apply it to a true indeterminate form — never to a determinate one. Series expansion ($e^x=1+x+\tfrac{x^2}{2}+\cdots$, $\sin x=x-\tfrac{x^3}{6}+\cdots$) often does the same job faster. Faster still — the asymptotic reading of the standard limits: as $x\to0$, $\sin x\sim x$, $\tan x\sim x$, $1-\cos x\sim\tfrac{x^2}{2}$, $e^x-1\sim x$, $\ln(1+x)\sim x$. Replace each factor by its leading power and compare orders: $\dfrac{(1-\cos x)^2}{x^3\sin x}\sim\dfrac{x^4/4}{x^4}=\tfrac14$; $\dfrac{\sin 3x\,\tan 2x}{x^2}\to 6$; a numerator of higher order than the denominator gives 0, lower gives $\pm\infty$.

If $f(x)\to 1$ and $g(x)\to\infty$, then $\lim [f(x)]^{g(x)}=e^{\,\lim\, g(x)\,[f(x)-1]}$. Equivalently, $L=e^{\lim g\ln f}$ — take $\ln$, evaluate the resulting product, then exponentiate.