The limit of a function describes the value that $f(x)$ approaches as $x$ gets arbitrarily close to a point $a$. We write $\lim_{x\to a} f(x)=L$ and say that $f(x)$ tends to $L$ as $x$ tends to $a$. Importantly, the function does not have to be defined at $a$ itself: only the behavior of $f$ near $a$ matters. For example, $\frac{x-1}{x^2-6x+5}$ is undefined at $x=1$, yet its limit as $x\to 1$ exists. In YÖS problems limits appear in several standard forms: direct substitution, indeterminate forms such as $\frac{0}{0}$, limits at infinity, remarkable (special) limits, and continuity conditions with parameters. The left-hand limit $\lim_{x\to a^-} f(x)$ is the value approached when $x$ approaches $a$ through values smaller than $a$; the right-hand limit $\lim_{x\to a^+} f(x)$ uses values greater than $a$. The two-sided limit exists if and only if both one-sided limits exist and are equal: $$\lim_{x\to a} f(x)=L \iff \lim_{x\to a^-} f(x)=\lim_{x\to a^+} f(x)=L.$$ One-sided limits are essential for absolute values and piecewise functions. Classic example: $\lim_{x\to 0^+}\frac{|x|}{x}=1$ and $\lim_{x\to 0^-}\frac{|x|}{x}=-1$, so $\lim_{x\to 0}\frac{|x|}{x}$ does not exist.
If $\lim_{x\to a} f(x)=A$ and $\lim_{x\to a} g(x)=B$, then: $$\lim_{x\to a}\bigl(f(x)\pm g(x)\bigr)=A\pm B,\qquad \lim_{x\to a} f(x)\cdot g(x)=A\cdot B,$$ $$\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{A}{B}\ (B\neq 0),\qquad \lim_{x\to a}\bigl(f(x)\bigr)^n=A^n.$$ Polynomials and all elementary functions are continuous on their domains, so their limits are found by direct substitution. Example: $\lim_{x\to 2}(x^3-5x+1)=2^3-5\cdot 2+1=8-10+1=-1$. Always try substitution first; only if it gives an indeterminate form like $\frac{0}{0}$ or $\frac{\infty}{\infty}$ do you need special techniques.
If substituting $x=a$ gives $\frac{0}{0}$, then $(x-a)$ is a common factor of the numerator and the denominator. Factor both, cancel $(x-a)$, and substitute again. Useful factorizations: $x^2-b^2=(x-b)(x+b)$, quadratics via roots $x^2+px+q=(x-x_1)(x-x_2)$, and taking out a common factor. Example: $\lim_{x\to 1}\frac{x-1}{x^2-6x+5}=\lim_{x\to 1}\frac{x-1}{(x-1)(x-5)}=\lim_{x\to 1}\frac{1}{x-5}=-\frac{1}{4}$. Similarly, $3x^2-27=3(x-3)(x+3)$ prepares the fraction $\frac{3x^2-27}{\sin(x-3)}$ for the remarkable limit $\frac{\sin t}{t}\to 1$ with $t=x-3$.
Open the absolute value according to the side from which $x$ approaches the point: $|x-a|=x-a$ for $x>a$ and $|x-a|=-(x-a)$ for $x<a$. Then compute the one-sided limit as usual. Example: for $\lim_{x\to 3^+}\frac{|x-3|+3x-9}{x^2-3x-4}$, since $x>3$ we have $|x-3|=x-3$, so the numerator is $(x-3)+3(x-3)=4(x-3)\to 0$, while the denominator tends to $9-9-4=-4\neq 0$; hence the limit equals $\frac{0}{-4}=0$. If a problem asks for a two-sided limit with a modulus, compare the left and right results: if they differ (as in $\frac{|x|}{x}$ at $0$), the limit does not exist.
The key special limits as $x\to 0$: $$\lim_{x\to 0}\frac{\sin x}{x}=1,\qquad \lim_{x\to 0}\frac{\tan x}{x}=1,\qquad \lim_{x\to 0}\frac{1-\cos x}{x^2}=\frac{1}{2},$$ $$\lim_{x\to 0}\frac{e^x-1}{x}=1,\qquad \lim_{x\to 0}\frac{a^x-1}{x}=\ln a.$$ The scaling rule: $\lim_{x\to 0}\frac{\sin kx}{x}=k$ and $\lim_{x\to 0}\frac{\tan kx}{x}=k$, since $\frac{\sin kx}{x}=k\cdot\frac{\sin kx}{kx}$. Powers work too: $\lim_{x\to 0}\frac{\sin^3 2x}{x^3}=\left(\lim_{x\to 0}\frac{\sin 2x}{x}\right)^3=2^3=8$. For $x\to\infty$ expressions like $2x\sin\frac{3}{4x}$, substitute $t=\frac{3}{4x}\to 0$: $2x\sin\frac{3}{4x}=\frac{3}{2}\cdot\frac{\sin t}{t}\to\frac{3}{2}$.
The second remarkable limit: $$\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x=e,\qquad \lim_{x\to\infty}\left(1+\frac{k}{x}\right)^x=e^k,\qquad \lim_{t\to 0}(1+t)^{1/t}=e.$$ It resolves the indeterminate form $1^{\infty}$. Method for $\lim\left(\frac{P(x)}{Q(x)}\right)^{g(x)}$ where the base tends to $1$: write the base as $1+\alpha(x)$ with $\alpha(x)\to 0$; then the limit equals $e^{L}$ where $L=\lim \alpha(x)\cdot g(x)$. Example: $\frac{4x+3}{4x+5}=1+\frac{-2}{4x+5}$, so $$\lim_{x\to\infty}\left(\frac{4x+3}{4x+5}\right)^{12x+3}=e^{\lim \frac{-2(12x+3)}{4x+5}}=e^{-6}.$$
As $x\to\pm\infty$, only the leading (highest-degree) terms of the numerator and denominator matter. For $\frac{a_n x^n+\dots}{b_m x^m+\dots}$: if $n<m$ the limit is $0$; if $n=m$ the limit is $\frac{a_n}{b_m}$; if $n>m$ the limit is infinite. Parameter problems exploit this. Example: find $a$ such that $\lim_{x\to-\infty}\frac{(a-2)x^2+x+1}{-3x+1}=-\frac{1}{3}$. A finite nonzero limit requires equal degrees, so the $x^2$ term must vanish: $a-2=0$, i.e. $a=2$. Then the expression becomes $\frac{x+1}{-3x+1}$ with limit $\frac{1}{-3}=-\frac{1}{3}$, as required.
The form $\infty-\infty$ with square roots is resolved by multiplying and dividing by the conjugate: $\sqrt{A}-\sqrt{B}=\frac{A-B}{\sqrt{A}+\sqrt{B}}$. Example: $$\lim_{x\to\infty}\left(\sqrt{4x^2+2x-5}-\sqrt{4x^2+5x+2}\right)=\lim_{x\to\infty}\frac{(4x^2+2x-5)-(4x^2+5x+2)}{\sqrt{4x^2+2x-5}+\sqrt{4x^2+5x+2}}=\lim_{x\to\infty}\frac{-3x-7}{2x+2x}=-\frac{3}{4},$$ since each root behaves like $2x$ for large $x>0$. Useful shortcut for $x\to+\infty$: $$\sqrt{ax^2+bx+c}-\sqrt{ax^2+b_1x+c_1}\to\frac{b-b_1}{2\sqrt{a}}.$$ Here $\frac{2-5}{2\cdot 2}=-\frac{3}{4}$.
Key facts: for $a>1$, $a^x\to\infty$ as $x\to+\infty$ and $a^x\to 0$ as $x\to-\infty$; also $\left(\frac{a}{b}\right)^x\to 0$ as $x\to-\infty$ if $\frac{a}{b}>1$. Strategy: divide numerator and denominator by the term that does NOT vanish, then use $q^x\to 0$. Example: $\lim_{x\to-\infty}\frac{5^x-3^{x+2}}{5^x+3^{x-2}}$. Rewrite $3^{x+2}=9\cdot 3^x$ and $3^{x-2}=\frac{1}{9}\cdot 3^x$, then divide by $3^x$: $$\lim_{x\to-\infty}\frac{\left(\frac{5}{3}\right)^x-9}{\left(\frac{5}{3}\right)^x+\frac{1}{9}}=\frac{0-9}{0+\frac{1}{9}}=-81,$$ since $\left(\frac{5}{3}\right)^x\to 0$ as $x\to-\infty$. As $x\to+\infty$ you would instead divide by the largest base, $5^x$.
A function $f$ is continuous at $x=a$ if three conditions hold: (1) $f(a)$ is defined; (2) $\lim_{x\to a} f(x)$ exists, i.e. $\lim_{x\to a^-} f(x)=\lim_{x\to a^+} f(x)$; (3) $\lim_{x\to a} f(x)=f(a)$. For a piecewise function the only suspicious points are where the formula changes. Parameter problems: write the equations $$\lim_{x\to a^-} f(x)=\lim_{x\to a^+} f(x)=f(a)$$ and solve the resulting system for the parameters (for instance $m$ and $n$), then compute the requested quantity such as $m\cdot n$. Elementary functions (polynomials, $\sin$, $\cos$, $e^x$, roots on their domains) are continuous everywhere they are defined, which justifies direct substitution.
If substitution gives $\frac{c}{0}$ with $c\neq 0$, the limit is not finite: the function grows without bound near the point, and the sign must be checked on each side. Example: $\lim_{x\to 0}\frac{x-2}{x^5}$. The numerator tends to $-2\neq 0$; the denominator $x^5\to 0$ and changes sign at $0$. For $x\to 0^+$: $x^5>0$, so the fraction tends to $-\infty$; for $x\to 0^-$: $x^5<0$, so it tends to $+\infty$. The one-sided limits are infinite and different, so the limit does not exist (the function diverges). In test answers such cases are described as "the limit does not exist" or "tends to infinity", depending on whether the two sides agree.
Example 1. $\lim_{x\to 2}(x^3-5x+1)$. A polynomial is continuous, so substitute: $8-10+1=-1$. Example 2. $\lim_{x\to 1}\frac{x-1}{x^2-6x+5}$. Substitution gives $\frac{0}{0}$. Factor: $x^2-6x+5=(x-1)(x-5)$, cancel $(x-1)$: $\lim_{x\to 1}\frac{1}{x-5}=\frac{1}{1-5}=-\frac{1}{4}$. Example 3. $\lim_{x\to 3}\frac{3x^2-27}{\sin(x-3)}$. Again $\frac{0}{0}$. Factor the numerator: $3x^2-27=3(x-3)(x+3)$, and let $t=x-3\to 0$: $$\lim_{x\to 3}\frac{3(x-3)(x+3)}{\sin(x-3)}=\lim_{t\to 0}\frac{t}{\sin t}\cdot 3(x+3)=1\cdot 3\cdot 6=18.$$
Example 1. $\lim_{x\to 0}\frac{\tan 6x+\sin 8x}{2x}$. Split the fraction: $\frac{\tan 6x}{2x}+\frac{\sin 8x}{2x}\to\frac{6}{2}+\frac{8}{2}=3+4=7$. Example 2. $\lim_{x\to 0}\frac{x\cos x}{e^{2x}-1}$. Write it as $\cos x\cdot\frac{2x}{e^{2x}-1}\cdot\frac{1}{2}$; since $\frac{e^{2x}-1}{2x}\to 1$ and $\cos x\to 1$, the limit is $\frac{1}{2}$. Example 3. $\lim_{x\to 0}\frac{1-\cos x}{x^2}=\frac{1}{2}$ directly by the standard formula. Example 4. $\lim_{x\to\infty}\left(\frac{4x+3}{4x+5}\right)^{12x+3}$: base $=1+\frac{-2}{4x+5}$, exponent product $\frac{-2(12x+3)}{4x+5}\to\frac{-24x}{4x}=-6$, so the answer is $e^{-6}$.
Let $$f(x)=\begin{cases} mx+n, & x<2 \\ 3, & x=2 \\ nx^2-m, & x>2 \end{cases}$$ and suppose $\lim_{x\to 2} f(x)=f(2)$. Find $m\cdot n$. Solution. Continuity at $x=2$ requires the left limit, the right limit and the value to coincide: left limit $\lim_{x\to 2^-}(mx+n)=2m+n$; right limit $\lim_{x\to 2^+}(nx^2-m)=4n-m$; value $f(2)=3$. The system $2m+n=3$ and $4n-m=3$ gives $n=3-2m$, so $4(3-2m)-m=3$, i.e. $12-9m=3$, hence $m=1$ and $n=1$. Therefore $m\cdot n=1$. This scheme — equate the one-sided limits to the function value and solve for the parameters — works for every continuity problem of this type.
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