Lecture 10: Finding Limits

10.1 Common Methods for Finding Limits

Common Methods: Eight Types

10.1.1 Method 1: Use the basic limit to find the limit

Common Basic Limits

$\lim_{x\to0}\frac{\sin x}x=1;$

Power Exponents and Square Roots

$\lim_{x\to0}(1+x)^{\frac1x}=e;$
is of type $1^\infty$, and the power base inside the parentheses is greater than 0;
$\lim_{x\to0}(1+\frac{1}{x})^{x}$ does not exist, because when x approaches 0, it can also tend toward 0 negative;
$\lim_{x\to0^+}(1+\frac{1}{x})^{x}=1$
$\lim_{x\to\infty}(1+\frac1x)^x=e;$
is of type $1^\infty$, and the base of the parentheses is greater than 0
$\lim_{x\to\infty}(1+x)^{\frac{1}{x}}$
$\lim_{x\to+\infty}(1+x)^{\frac{1}{x}}=1$
$\lim_{x\to0}\frac{a^x-1}x=\ln a;$
$\lim_{n\to\infty}\sqrt[n]{n}=1.$
Note that here it is n, so it tends toward positive infinity;
$\lim_{n\to\infty}\sqrt[n]{a}=1,(a>0).$
Note that here it is n, so it tends toward positive infinity;

Piecewise Functions

$\left.\lim_{{x\to\infty}}\frac{a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0}{{b}_mx^m+b_{m-1}x^{m-1}+\cdots+b_1x+b_0}=\left|\begin{array}{ll}\frac{a_n}{b_m},&n=m,\\0,&n<m,\\\infty,&n>m.\end{array}\right.\right.$
With so many numerator and denominator, the result mainly depends on the largest number (highest degree);
Note that x tends toward infinity.
When x approaches 0, the value of the function should depend on the smallest number (minimum degree);
$\lim_{{n\to\infty}}x^n=\begin{cases}0,&|x|<1,\\1,&x=1,\\+\infty,&x>1,\\\text{does not exist},&x\leq-1.\end{cases}$
$\lim_{n\to\infty}e^{{nx}}=\begin{cases}0,&x<0,\\+\infty,&x>0\\1,&x=0.&\end{cases}$

1 to the power of infinity

For $1^{\infty}$, its result is infinitive, which does not necessarily equal 1;
$\text{ If }\lim\alpha(x)=0,\lim\beta(x)=\infty,\text{ and }\lim\alpha(x)\beta(x)=A.\text{ then }\lim(1+\alpha(x))^{\beta(x)}=e^A$
That is, the core is at $\lim\alpha(x)\beta(x)=A$, check if it holds;
The method trilogy
1. Written in standard form: $\text{ Original form }=\mathbf{lim}[1+\alpha(x)]^{\beta(x)}$;
2. Find the limit: $\lim\alpha(x)\beta(x)=A;$
3. Write the result: $\text{ Original form }=e^A.$
Example problem
Title: $\lim_{x\to\infty}\left(\frac{x^2}{(x-a)(x+b)}\right)^x$
$\lim_{x\to\infty}\left(\frac{x^2}{(x-a)(x+b)}\right)^x=\lim_{x\to\infty}\left(\frac{x_.}{x-a}\right)^x\left(\frac x{x+b}\right)^x=e^a\cdot e^{-b}=e^{a-b}$

Problem Type: Problem-solving technique - arbitrary constants

PART 1: Questions

When a multiple-choice question containsparameters that can take any value, consider giving the constant a special specific value. Substituting it can eliminate certain expressions without changing the original meaning;

PART 2: Typical Example Problems

In $\lim_{x\to\infty}\left(\frac{x^2}{(x-a)(x+b)}\right)^x$, a can be set to 0;

10.1.2 Method 2: Use equivalent infinitesimal substitution to find limits

Principle of Equivalent Substitution

1. Multiplication and division relationships can be swapped
$\text{ If }\alpha\sim\alpha_1,\beta\sim\beta_1,\text{ then }\lim\frac\alpha\beta=\lim\frac{\alpha_1}\beta=\mathbf{lim}\frac\alpha{\beta_1}=\mathbf{lim}\frac{\alpha_1}{\beta_1}$
2. Addition and subtraction relationships can be swapped undercertain conditions
Subtraction
$\text{ If }\alpha{\sim}\alpha_1,\beta{\sim}\beta_1,\text{ and }\lim\frac{\alpha_1}{\beta_1}=A\neq1.\text{ then }\alpha{-}\beta{\sim}\alpha_1{-}\beta_1$
Conditions for subtractive exchange:The two subtraction terms are not equal, A is not equal to 1;
For example: $sin2x - tanx$ ~ $2x - x$
Addition
$\text{ If }\alpha{\sim}\alpha_1,\beta{\sim}\beta_1,\text{ and }\lim\frac{\alpha_1}{\beta_1}=A\neq-1.\text{ then }\alpha{-}\beta{\sim}\alpha_1{+}\beta_1$
Subtractive substitution condition: Two additive terms are not equal, A is not equal to -1;

Common Equivalent Infinitesimals

When x approaches 0;
${(1+x)^\alpha-1}\sim{\alpha x}$
$e^{x}-1\sim x$
Power refers to the substitution of a function;
$x\sim\sin x\sim\tan x\sim\arcsin x\sim\arctan x\sim\ln(1+x)\thicksim e^x-1;$
Note: For this function, you need to ensure that x tends toward 0 when using it;
Variant: $\ln q(x)\sim q(x)-1,q(x)\to1$
$a^x-1\sim x\ln a,\quad{(1+x)^\alpha-1\sim\alpha x},\quad1-\cos x\sim\frac12x^2$

Advanced Equivalent Infinitesimal

$x-\sin x\sim\frac16x^3$，$\tan x-x\sim\frac13x^3$
$\arcsin x-x\sim\frac16x^3\quad x-\arctan x\sim\frac13x^3$
$x-\ln(1+x)\sim\frac12x^2$

10.1.3 Method 3: Use rational algorithms to calculate limits

What is the Rational Operation Rule

The addition, subtraction, multiplication, and division of limits can be calculated separately, providedthe limit of each split part must exist;

\begin{aligned} \text{If }{\lim f(x)=A}, \quad\mathrm{lim~g(x)=B}, \text{ Then}: &\lim(f(x)\pm g(x))=\lim f(x)\pm\lim g(x) \\ &\lim(f(x)\cdot g(x))=\lim f(x)\cdot\lim g(x) \\ &\lim\left(\frac{f(x)}{g(x)}\right)=\frac{\lim f(x)}{\lim g(x)}\quad(B\neq0) \end{aligned}

Prerequisites for Existence

1. Existence +/- Nonexistence = Nonexistence;
2. Not existent +/- Not existent = Not necessarily;
3. Exist ×/÷ Not exist = Not necessarily;
Can exist, can not exist;
4. Not existing ×/÷ Not existing = Not necessarily so;
For example: $(-1)^{n}\cdot(-1)^{n}=1$, when two of them are individual, they do not exist, but when multiplied by combination, they exist;

Rational Operation Rules: Common Conclusions

1. $\lim f(x)=A\neq0\Rightarrow\lim f(x)g(x)=A\lim g(x);$
The limit of a nonzero factor can be determined first;
2. $\lim\frac{f(x)}{g(x)}$ exists, $\lim g(x)=0\Rightarrow\lim f(x)=0; $
3. $\lim\frac{f(x)}{g(x)}=A\neq0,\lim f(x)=0\Rightarrow\lim g(x)=0;$

10.1.4 Method 4: Use L'Hôpital's Rule to Find Limits

Usage Requirements

1. $\lim_{x\to x_0}f(x)=\lim_{x\to x_0}g(x)=0\left(\infty\right);$
2. $f$ and $g$ are differentiable in a punctured neighborhood of $x_0$, and $g'(x)\neq0$.
L'Hôpital's rule follows from Cauchy's mean value theorem, so these hypotheses are essential.
3. $\lim_{x\to x_0}\frac{f^{\prime}(x)}{g^{\prime}(x)}$ exists (or );
Conditions met: $\lim_{x\to x_0}\frac{f(x)}{g(x)}=\lim_{x\to x_0}\frac{f^{\prime}(x)}{g^{\prime}(x)}.$
Note: To determine whether this equality holds true, you need to check whether it exists on the right side of the equation;

Usage Environment

$\frac00;\quad\frac\infty\infty,\quad0\cdot\infty;\quad\infty-\infty;\quad1^\infty;\quad\infty^0;\quad0^0$
In fact, they are all derived from $\frac00;\quad\frac\infty\infty$:
Problem-solving approach:
$\frac00,\frac\infty\infty\quad\Leftarrow\begin{cases}\quad ∞-∞\\0\cdot\infty\quad\Leftarrow\begin{cases}1^\infty\\\infty^0\\0^0&\end{cases}&\end{cases}$
$0\cdot\infty\quad$
Approach: Place either 0 or infinity on top OR below;
$1^\infty;\quad\infty^0;\quad0^0$：
Approach: All three should be rewritten as 0 times infinity;
Method: $[f(x)]^{g(x)}=e^{g(x)\ln f(x)}$; rewrite the expression as an exponential.
Core: Find the limit of the exponent $g(x)\ln f(x)$.

Summary

1. If the assumptions guarantee only two derivatives of $f$, L'Hôpital's rule can be applied at most twice, provided all of its hypotheses remain satisfied.
2. N-order differentiable -> n-1 order differentiable can be used;
3. If you encounter abstract functions to find limits, you can consider using the definition of derivatives for calculation;

10.1.5 Method 5: Use Taylor's Formula to Find Limits

#####Theorem: Taylor's Theorem—with the remaining Peano terms

description:

1. Taylor's formula with Peano remainder: Suppose $f(x)$ is differentiable of $n$ order at $x_0$, then $f(x)=\sum_{k=0}^n\frac{f^{(k)}(x_0)}{k!}(x-x_0)^k+o((x-x_0)^n)$

Meaning: Write a general function as a general polynomial + remainder

Where: $$P_n(x)=\sum_{k=0}^n\frac{f^{(k)}(x_0)}{k!}(x-x_0)^k$$

$f(x)$ n Taylor polynomials at $x_{0}$

$$R_n(x)=o((x-x_0)^n)$$

Disadvantages:

> 1. Only qualitative descriptions of the remaining items can be given, not quantitative analysis;

2. It can only be used when x0 is approaching; it cannot be used from a distance;

Common Taylor Formulas

$$\mathrm{e^x=\sum_{n=0}^\infty\frac1{n!}x^n=1+x+\frac1{2!}x^2+\cdots\in(-\infty,+\infty)}$$
$$\sin\mathrm{x=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}x^{2n+1}=x-\frac1{3!}x^3+\frac1{5!}x^5+\cdots,x\in(-\infty,+\infty)}$$
$$\cos\mathrm{x=\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}x^{2n}=1-\frac1{2!}x^2+\frac1{4!}x^4+\cdots,x\in(-\infty,+\infty)}$$
$$\mathrm{ln(1+x)=\sum_{n=0}^\infty\frac{(-1)^n}{n+1}x^{n+1}=x-\frac12x^2+\frac13x^3+\cdots,x\in(-1,1]}$$
$$\begin{aligned}&\frac1{1-\mathrm{x}}=\sum_{\mathrm{n=0}}^\infty\mathrm{x^n}=1+\mathrm{x+x^2+x^3+\cdots,x\in(-1,1)}\\&\frac1{1+\mathrm{x}}=\sum_{\mathrm{n=0}}^\infty(-1)^\mathrm{n}\mathrm{x}^\mathrm{n}=1-\mathrm{x}+\mathrm{x}^2-\mathrm{x}^3+\cdots,\mathrm{x}\in(-1,1)\end{aligned}$$
$$\mathrm{(1+x)^a=1+\sum_{n=1}^\infty\frac{a(a-1)\cdots(a-n+1)}{n!}x^n=1+ax+\frac{a(a-1)}{2!}x^2+\cdots,x\in(-1,1)}$$

$\frac{0}{0}$ Common Methods

Equivalent substitution;
L'Hôpital's rule;
Taylor's formula;

How many times is Taylor's Formula used

1. $\frac{f(x)}{g(x)}.$
How to use:
1. Using Taylor expansion, the expansion has the same power index above and below the order;
For example: the top and bottom are to the fourth power, and the bottom and top are also expanded to the fourth power;
2. Organize terms, organize terms of the same order together;
2. $f(x)-\text{ or }+g(x)$
Write about a power exponent of a certain degree and find the lowest degree term to subtract them;

10.1.6 Method 6: Use the squeeze principle to find limits

Example Question: $\lim_{n\to\infty}\left[\frac1{n^2+n+1}+\frac2{n^2+n+2}+\cdots+\frac n{n^2+n+n}\right]$

Analysis
Reduction: All denominators are reduced to the largest denominator;
Enlargement: Convert all denominators to the smallest denominator;
Analysis
$\frac{\frac{1}{2}n(n+1)}{n^{2}+n+n}\leq[\text{ Original form }]\leq\frac{\frac{1}{2}n(n+1)}{n^{2}+n+1}$
Both before and after tend to be 1/2, so the original expression = 1/2;
Question Type: Principleofsqueezing

Example Question: $\lim_{n\to\infty}\sqrt[n]{a_1^n+{a_2}^n+\cdots+{a_m}^n},\quad\text{ where }a_i>0,(i=1,2,\cdots m)$

Analysis
Shrink the largest term to: $\sqrt[u]{ma^{n}}$ the largest term among all current terms
How to shrink the smallest term to a result that is also a? Keep only one $a^n$ item: $\sqrt[n]{a^{n}}$
Analysis
In the largest term, since m is a constant, it ultimately converges to a;
$\sqrt[n]{a^{n}}\leq\sqrt[n]{a_{1}^{n}+a_{1}^{n}+\cdots+a_{m}^{n}}\leq\sqrt[n]{ma_{0}^{n}}$
Question Type: Principleofsqueezing

Method: When encountering a finite term in the square root, just find the term that most affects the current total number size, with its limit being the scaling value of the squeeze principle;

Example: $\lim_{n\to\infty}\sqrt[n]{1^n+x^n+(\frac{x^2}2)^n},{(x>0).}$
There are three of these in total. Find the one with the largest base: $\max\{1,x,\frac{x^2}2\}$
When x approaches infinity, $\frac{x^2}2$ has the greatest influence;

10.1.7 Method 7: Use the monotone bounded criterion to find limits

Method

Generally divided into two steps:
1. Proving the existence of -> monotonic bounded criterion;
2. Take the limits on both sides of the equation;

Example Question: $\text{ Let }x_1>0,x_{n+1}=\frac12\Bigg(x_n+\frac1{x_n}\Bigg),n=1,2,\cdots.\text{ Find the limit }\lim_{n\to\infty}x_n$

Analysis:
Inequality conclusion: $2ab\leq a^{2}+b^{2}$
Analysis
Step 1: Prove the existence of limits
Since x 1>0, $x_n>0$;
$x_{n+1}=\frac12\left(x_n+\frac1{x_n}\right)=\frac12\left[(\sqrt{x_n})^2+(\frac1{\sqrt{x_n}})^2\right]\geq\frac12\cdot2\sqrt{x_n}\cdot\frac1{\sqrt{x_n}}=1$
$x_{n+1}-x_{n}=\frac{1}{2}\Bigg(\frac{1}{x_{n}}-x_{n}\Bigg)=\frac{1}{2}\cdot\frac{1-x_{n}^{2}}{x_{n}}\leq0$
Because $x_{n+1}-x_{n}<0$ is monotonically decreasing and the function has a lower bound, so the limit exists;
Step 2: Take the limits on both sides of the equation and find the limit
$\lim_{n\to\infty}x_{n}=a.$
$a=\frac12{\left(a+\frac1a\right)}$
$\lim_{n\to\infty}x_n=\mathbf{1}.$
Question Type: Monotonyandboundedcriteria

Concept: Monotonic bounded criterion used in various scenarios

Mostly used to find recurrence relations $x_{n+1}=f(x_n)$

10.1.8 Method 8: Use definite integrals to find limits

Concept of Definite Integrals

Definite integrals sum all terms within an interval (for example, from a to b);
Intervals are divided and taken according to patterns;
N equal divisions between a and b -> Each part is $\frac{b-a}{n}$;

Example Question: $\text{ Find the limit }\lim_{n\to\infty}\left[\frac1{n+1}+\frac1{n+2}+\cdots+\frac1{n+n}\right]$

Analysis
This question is very difficult to extract using the squeezing principle;
Definite integrals are also "Japanese limits"
Analysis
Step 1: Extract 1/n (Cute Factor~)
Original form: $=\lim_{n\to\infty}{\frac1n}{\left[\frac1{1+\frac1n}+\frac1{1+\frac2n}+\cdots+\frac1{1+\frac nn}\right]}$;
Step 2: Find the changing part -> Which is the integrand
The changing part is the position of the function x, and the range of change is the range of finding definite integrals (from $\frac1n$ to n, where n values range from 0 to 1);
Thus, the changed part becomes x: $\int_{0}^{1}\frac{1}{1+x}dx$
1. Where there is change, there is variable;
2. The range of change in X is the range of the interval;
Question Type: Definiteintegralsdefinethelimittofindthelimit

Summary:n items and questions

There are two solutions:
1. Principle of compression;
2. Definition of definite integrals;