Limit Cheat Sheet: Master Calculus Limits Quickly

A Limit Cheat Sheet is a concise guide that helps students quickly recall important limit concepts, rules, and techniques used in calculus. It typically covers topics like the definition of a limit, L'Hopital's Rule, one-sided limits, and common limit-solving strategies. This quick-reference tool is perfect for exam preparation or reinforcing fundamental limit calculations in calculus.

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Limits Definitions

Precise Definition : We say ${\lim\limits_{x \to a} f (x) = }$ if for every ${\varepsilon>0}$ there is a ${\delta>0}$ such that whenever ${0<|x-a|<\delta}$ then ${f(x)-L|<\varepsilon.}$

Definition : We say ${\lim\limits_{x \to a} f (x) = L}$ if we can make ${f(x)}$ as close to ${L}$ as we want by taking ${x}$ sufficiently close to ${a}$ (on either side of ${a}$) with out letting ${x=a.}$

Right-hand limit: ${\lim\limits_{x \to a^+} f (x) = L.}$ This has the same definition as the limit except it requires ${x<a.}$

Left-hand limit: ${\lim\limits_{x \to a^-} f (x) = L.}$ This has the same definition as the limit, except it requires ${x<a.}$

Limit at infinity : We say ${\lim\limits_{x \to \infty} f (x) = L}$ if we can make ${f(x)}$ as close to ${L}$ as we want by taking ${x}$ large enough and positive.

There is a similar definition for ${\lim\limits_{x \to -\infty} f (x) = L}$ except we require ${x}$ large and negative.

Infinite limit : We say ${\lim\limits_{x \to a} f (x) = \infty}$ if we can make ${f(x)}$ arbitrarily large (and positive) by taking ${x}$ sufficiently close to ${a}$ (on either side of ${a}$ without letting ${x=a.}$

There is a similar definition for ${\lim\limits_{x \to a} f (x) = - \infty}$ except we make ${f(x)}$ arbitrarily large and negative.

Relationship between the limit and one-sided limits:

• $\lim\limits_{x \to a} f (x) = L \rArr \lim\limits_{x \to a^+} f (x) = \lim\limits_{x \to a^-} f (x) = L$

  
  

• $\lim\limits_{x \to a^+} f (x) \neq \lim\limits_{x \to a^-} f (x) \rArr \lim\limits_{x \to a} f (x) \rArr$ It means Limit Does Not Exist

Properties of Limits:

Assume $\lim\limits_{x \to a}{f(x)}$ and $\lim\limits_{x \to a}{g(x)}$ both exist and $c$ is any number then,

1. $\lim\limits_{x \to a}{[cf(x)]} = c \lim\limits_{x \to a}{f(x)}$

   
   

2. $\lim\limits_{x \to a}{[f(x)\pm {g(x)}]} = \lim\limits_{x \to a}{f(x)\pm {\lim\limits_{x \to a} g(x)}}$

   
   

3. $\lim\limits_{x \to a}{[f(x){g(x)}]} = \lim\limits_{x \to a}{f(x) \space \lim\limits_{x \to a} {g(x)}}$

   
   

4. $\lim\limits_{x \to a} {\bigg[\dfrac{f(x)}{g(x)}\bigg]} = \dfrac{\lim\limits_{x \to a}{f(x)}}{\lim\limits_{x \to a}{g(x)}}$ provided ${\lim\limits_{x \to a}{g(x)} \neq 0}$

   
   

5. $\lim\limits_{x \to a}\big[f(x)]^n = \bigg[\lim\limits_{x \to a}[f(x)\bigg]^n$

   
   

6. $\lim\limits_{x \to a}{\bigg[\sqrt[n]{f(x)}\bigg] = \sqrt[n]{\lim\limits_{x \to a}f(x)}}$

   
   

Limit Evaluations at ${\pm}\infty$:

1. $\lim\limits_{x \to \infty}{e^x}= \infty \space \space \& \space \lim\limits_{x \to - \infty}{e^x} = 0$

   
   

2. $\lim\limits_{x \to \infty}in(x) = \infty \space \& \space \lim\limits_{x \to 0^+}in(x) = -\infty$

   
   

3. if ${r>0}$ then $\lim\limits_{x \to \infty} \dfrac{b}{x^r} = 0$

   
   

4. if ${r>0}$ and ${x^r}$ is real for negative $x$ then $\lim\limits_{x \to -\infty} \dfrac{b}{x^r} = 0$

   
   

5. $n$ even : $\lim\limits_{x \to \pm \infty}{x^n} = \infty$

   
   

6. $n$ odd : $\lim\limits_{x \to \infty}{x^n} = \infty \space \& \space \lim\limits_{x \to -\infty}{x^n} = -\infty$

   
   

7. $n$ even : $\lim\limits_{x \to \pm \infty}{ax^n}+ \dots + bx + c = sgn(a)\infty$

   
   

8. $n$ odd : $\lim\limits_{x \to \infty}{ax^n}+ \dots + bx + c = sgn(a)\infty$

   
   

9. $n$ odd : $\lim\limits_{x \to - \infty}{ax^n}+ \dots + cx + d = -sgn(a)\infty$

   
   

Note : ${sgn(a)=1}$ if ${a>0}$ and ${sgn(a) = -1}$ if ${a<0.}$

Limit Evaluation Techniques:

Continuous Function

If ${f(x)}$ is Continuous at $a$ then $\lim\limits_{x \to a}{f(x)=f(a)}$

Continuous Functions and Composition

${f(x)}$ is Continuous at $b$ and $\lim\limits_{x \to a}{g(x)=(b)}$ then

$\lim\limits_{x \to a}{f(g(x)_=f} \bigg(\lim\limits_{x \to a}{g(x)\bigg)=f}(b)$

Factor and Cancel

$\space \lim\limits_{x \to 2}\dfrac{x^2-x-2}{x^2-2x} = \lim\limits_{x \to 2}\dfrac{(x-2)(x+1)}{x(x-2)} =\lim\limits_{x \to 2}\dfrac{x+1}{x}= \dfrac{3}{2}$

Rationalize Numerator/Denominator

$\space \lim\limits_{x \to 9} \dfrac{3-\sqrt x}{x^2-81} = \lim\limits_{x \to 9} \dfrac{3-\sqrt x}{x^2-81} \space\dfrac{3+\sqrt x}{3+\sqrt x} = \lim\limits_{x \to 9} \dfrac{9-x}{(x^2-81)(3+\sqrt x)} = \lim\limits_{x \to 9} \dfrac{-1}{(x+9)(3+\sqrt x)} = - \dfrac{1}{108}$

Combine Rational Expressions

$\space \lim\limits_{h \to 0} \dfrac{1}{h} \bigg(\dfrac{h}{x+h}-\dfrac{1}{x}\bigg) = \lim\limits_{h \to 0} \dfrac{1}{h} {\bigg(\dfrac{x-(x+h)}{x(x+h)}\bigg)} = \lim\limits_{h \to 0} \dfrac{1}{h} \bigg(\dfrac{-h}{x(x+h)}\bigg) = \lim\limits_{h \to 0} \bigg(\dfrac{-1)}{x(x+h)}\bigg) = -\dfrac{1}{x^2}$

L Hospital's or L Hopital's Rule

if $\space \lim\limits_{x \to a} \dfrac{f(x)}{g(x)} = \dfrac{0}{0}$ or $\lim\limits_{x \to a} \dfrac{f(x)}{g(x)} = \dfrac{\pm \infty}{\pm \infty}$ then,

$\space \lim\limits_{x \to a} \dfrac{f(x)}{g(x)} = \lim\limits_{x \to a} \dfrac{f'(x)}{g'(x)}, a$ is a number, $\infty$ or ${- \infty}$

Polynomials at Infinity

${p(x)}$ and ${q(x)}$ are polynomials. To compute

$\lim\limits_{x \to a \pm \infty} \dfrac{p(x)}{q(x)}$ factor largest Power of $x$ in ${q(x)}$ out of

both ${p(x)}$ and ${q(x)}$ then compute limit.

$\lim\limits_{x \to - \infty} \dfrac{3x^2-4}{5x-2x^2} = \lim\limits_{x \to - \infty} \dfrac{x^2(3-\frac{4}{x^2})}{x^2(\frac{5}{x}{-2})} = \lim\limits_{x \to - \infty} \dfrac{3-\frac{4}{x^2}}{\frac{5}{x}{-2}} = -\dfrac{3}{2}$

Piecewise Function

$\lim\limits_{x \to - 2}{g(x)}$ where ${g(x)} = \begin{cases} {x^2+5} &\text{if }{x<-2} \\ {1-3x} &\text{if } {x \geq -2} \end{cases}$

Compute two one sided limits,

$\lim\limits_{x \to - 2^-}{g(x)} = \lim\limits_{x \to - 2^-}{x^2+5=9}$

$\lim\limits_{x \to - 2^+}{g(x)} = \lim\limits_{x \to - 2^+}{1-3x=7}$

One sided limits are different so $\lim\limits_{x \to - 2}{g(x)}$ dosen't
exist. If the two one sided limits had been equal
then $\lim\limits_{x \to - 2}{g(x)}$ would have existed and had the
same value.

Continuous Function Examples:

Partial list of continuous functions and the values of $x$ for which they are continuous.

1. Polynomials for all $x$.

   
   

2. Rational function, except for $x'$s that give division by Zero.

   
   

3. ${\sqrt[n]x}$ (n odd) for all $x.$

   
   

4. ${\sqrt[n]x}$ (n even) for all ${x \geq 0}.$

   
   

5. ${e^x}$ for all $x.$

   
   

6. In $(x)$ for ${x>0}.$

   
   

7. ${\cos(x)}$ and ${\sin(x)}$ for all $x.$

   
   

8. ${\tan(x)}$ and ${\sec(x)}$ provided
   $x \neq \dots, - \dfrac{3\pi}{2}, \dfrac{\pi}{2}, \dfrac{\pi}{2}, \dfrac{3\pi}{2}, \dots$

   
   

9. ${\cot(x)}$ and ${\csc(x)}$ provided
   $x \neq \dots, {-2\pi}, {-\pi}, 0, {\pi}, {2\pi}, \dots$

   
   

Intermediate Value Theorem:

Suppose that ${f(x)}$ is continuous on ${[a,b]}$ and let $M$ be any number between ${f(a)}$ and ${f(b)}.$

Then there exists a number $c$ such that ${a<c<b}$ and ${f(c))= M.}$

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