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https://github.com/quantum-software-development/limits-calculus-i

✍️ Math Calculus I - Limits and Derivatives Exercises - AI Data Science - PUCSP University
https://github.com/quantum-software-development/limits-calculus-i

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✍️ Math Calculus I - Limits and Derivatives Exercises - AI Data Science - PUCSP University

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 ## 
 ✍️ Limits - Calculus I - Resolution of Mathematics Exercises








 










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#### 
 AI Data Science - 1st Semester / 20024 - PUCSP University - Math Repository - [Professor Eric Bacconi Gonçalves](https://www.linkedin.com/in/eric-bacconi-423137/)



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### 
 [![Sponsor Quantum Software Development](https://img.shields.io/badge/Sponsor-Quantum%20Software%20Development-brightgreen?logo=GitHub)](https://github.com/sponsors/Quantum-Software-Development)







## [1. Find the limits:]()



### 1a)  **Limit Expression:**



$$lim_{{x \to 3}} \frac{{x^2 - 9}}{{x - 3}}$$







**Simplified Form:** The numerator  $$\large x^2 - 9$$, can be factored as (x + 3)(x - 3) ), which simplifies the expression to:







$$\\begin{align*}

\large \lim_{{x \to 3}} (x + 3)


\end{align*}

\$$







[**Final Result:**]()



Substituting ( x ) with 3, we get: 



$$\large  3 + 3 = 6$$







**Explanation:** The limit as ( x ) approaches 3 for the function $\large \frac{{x^2 - 9}}{{x - 3}}$ is 6.



This is because the factor ( x - 3 ) in the denominator cancels out with the same factor in the numerator, leaving ( x + 3 ) which evaluates to 6 when ( x ) is 3.

 

#



### 1b) **The Limit Expression given is:**



$$\lim_{{x \to -7}} \frac{{49 - x^2}}{{7 + x}}$$







**Simplified Form:** The numerator $\large ( 49 - x^2 )$ is a difference of squares and can be factored as $\large (7 + x)(7 - x)$.







**This allows us to simplify the expression by canceling out the common factor of:**  $\large ( 7 + x )$ in the numerator and denominator:



$$\large \lim_{{x \to -7}} (7 - x)$$







[**Final Result:**]()



When we substitute ( x ) with -7, the expression simplifies to:
7 - (-7) = 14



 



**Explanation:** The limit of the function $(\Large \frac{{49 - x^2}}{{7 + x}} )$ as ( x ) approaches -7 is 14.



This result is obtained because after canceling the common factor, we are left with ( 7 - x ), which equals 14 when ( x ) is -7.



  #



### 1c) **Limit Expression:**



$$

\lim_{{x \to 1}} \frac{{x^2 - 4x + 3}}{{x - 1}}

$$



[Solution]()



To solve the limit, we can factor the numerator:



$$

x^2 - 4x + 3 = (x - 1)(x - 3)

$$



So the limit becomes:



$$

\lim_{{x \to 1}} \frac{{(x - 1)(x - 3)}}{{x - 1}}

$$



We can cancel out the \((x - 1)\) terms:



$$

\lim_{{x \to 1}} (x - 3)

$$



Now, we can directly substitute \( x = 1 \):



$$

1 - 3 = -2

$$



Therefore, the limit is:



$$

\lim_{{x \to 1}} \frac{{x^2 - 4x + 3}}{{x - 1}} = -2

$$



#



### 1d)   **Limit Expression:**




$$\lim_{{x \to 1}} \frac{{x^2 - 2x + 1}}{{x - 1}}$$







To calculate the limit, we can simplify the expression by factoring the numerator, which is a perfect square trinomial. Factoring (x^2 - 2x + 1), we get ((x - 1)(x - 1)). The denominator is already in factored form as (x - 1). Thus, the function simplifies to:



$$\frac{(x - 1)(x - 1)}{x - 1}$$



After canceling out the common term ( x - 1 ), we are left with:



$$\lim_{{x \to 1}} (x - 1)$$



Since there are no more terms that depend on ( x ), this simplifies to:



$$\lim_{{x \to 1}} = x - 1 = 0$$



[Final Result:]()



The limit of the function as ( x ) approaches 1 is simply 0.



#



### 1e)  **Limit Expression:**




$$\lim_{{x \to 2}} \frac{{x - 2}}{{x^2 - 4}}$$







To solve this limit, we need to factor the denominator and simplify the expression. The denominator ( x^2 - 4 ) can be factored into ( (x + 2)(x - 2) ), which allows us to cancel out the ( x - 2 ) term in the numerator:



$$\lim_{{x \to 2}} \frac{1}{{x + 2}}$$



Substituting ( x = 2 ) into the simplified expression, the final value:



$$\frac{1}{4}$$







[Final Result:]()



The limit of the function as ( x ) approaches 1 is simply $$\frac{1}{4}$$



#



### 1f) **Limit Expression:**




$$\(\lim_{{x \to 3}} \frac{{x^3 - 27}}{{x^2 - 5x + 6}}\)$$







This limit can be solved using factorization and polynomial division: 




$$\

\begin{align*}

\lim_{{x \to 3}} \frac{{x^3 - 27}}{{x^2 - 5x + 6}} 

&= \lim_{{x \to 3}} \frac{{(x - 3)(x^2 + 3x + 9)}}{{(x - 2)(x - 3)}} \\

&= \lim_{{x \to 3}} \frac{{x - 3}}{{x - 2}} \\

&= \frac{{3 - 3}}{{3 - 2}} \\

&= 1

\end{align*}

\$$







[Final Result:]()



The limit of the function as ( x ) approaches 1 is simply $$\frac{1}{4}$$



#



### 1g:  **Limit Expression:**




$$\(\lim_{{x \to \infty}} \frac{{x^2}}{{2x^2 - x}}\)$$







In this case, we can use L'Hôpital's rule, as the limit is of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) when \(x\) tends to infinity.



$$\

\begin{align*}

\lim_{{x \to \infty}} \frac{{x^2}}{{2x^2 - x}} &= \lim_{{x \to \infty}} \frac{{\frac{d}{dx}[x^2]}}{{\frac{d}{dx}[2x^2 - x]}} \\

&= \lim_{{x \to \infty}} \frac{{2x}}{{4x - 1}} \\

&= \lim_{{x \to \infty}} \frac{{2}}{{4 - \frac{1}{x}}} \\

&= \frac{2}{4} \\

&= \frac{1}{2}

\end{align*}

\$$



[Final Result:]() 



The limit of the expression is $$\frac{1}{2}$$







## [2. Solve the Limits:]()



### 2a):  **Limit Expression:**




 $\lim_{x \to \infty} \frac{1}{x^2}$ 




 



The limit as ( x ) approaches infinity for ( $\frac{1}{{x^2}}$ ):



is:
$\lim_{{x \to \infty}} \frac{1}{{x^2}} = 0$




As ( x ) increases without bound, the value of ( \frac{1}{{x^2}} ) approaches 0 because the denominator grows much faster than the numerator.



#



### 2b)  **Limit Expression:**




( $\lim_{x \to -\infty} \frac{1}{x^2}$ )







The limit as ( x ) approaches negative infinity for ( $\frac{1}{{x^2}}$ ) is:




 $\lim_{x \to -\infty} \frac{1}{x^2}$ = 0



As ( x ) decreases without bound, the value of ( $\frac{1}{{x^2}}$ ) approaches 0, similar to part a), because squaring a negative number results in a positive number, which grows larger.



#



### 2c)  **Limit Expression:**




$\lim_{x \to \infty} x^4$








The limit as ( x ) approaches infinity for ( x^4 ) is:
grows at an increasing rate and approaches infinity for ( x^4 ) is:



$\lim_{x \to \infty} x^4$
 = $\infty$



Similar to the previous expressions, the term ( 2x^5 ) grows at a faster rate than the others, causing the expression to approach infinity.



#



### 2d)  **Limit Expression:**




$\lim_{{x \to \infty}}$ $(2x^4 - 3x^3 + x + 6)$ 







The limit as ( x ) approaches negative infinity for ( 2x^4 - 3x^3 + x + 6 ) is:



$\lim_{x \to -\infty} (2x^4 - 3x^3 + x + 6) = \infty$




Even though ( x ) is negative, the highest power term ( x^4 )  will still lead the expression to increase without bound because the even power makes it positive.



#



### 2e)  **Limit Expression:**




2x^5 - 3x^2 + 6







The limit as ( x ) approaches infinity for $\( 2x^5 - 3x^2 + 6 ) is:




The limit as ( x ) approaches negative infinity for $\( 2x^4 - 3x^3 + x + 6 )$ is:




$\lim_{x \to -\infty} (2x^4 - 3x^3 + x + 6) = \infty$



Even though ( x ) is negative, y.







## [3.Calculate the Following Limits]()



### 3a: Finding the limit of a polynomial function as x approaches infinity



The given function is a polynomial function of the form: 



$$f(x)=axn+bxn−1+cxn−2+...+dx+e$$







As x approaches infinity, the highest power of x in the function dominates the value of the function. This means that we can ignore all the lower-order terms, and simply consider the behavior of the highest-order term.

In this case, the highest-order term is 2x4. As x approaches infinity, x4 also approaches infinity, and so the function f(x) also approaches infinity.



Therefore, the limit of the function as x approaches infinity is infinity. We can write this mathematically as:



$$x→∞lim x32x4−3x3+x+6 =0$$



#



### 3b:Finding the limit of a rational function as x approaches infinity



The given function is a rational function of the form 



$$f(x)=cxm+fxm−1+...+gx+haxn+bxn−1+...+dx+e$$







, where n > m. As x approaches infinity, the highest power of x in the numerator dominates the value of the numerator, and the highest power of x in the denominator dominates the value of the denominator.  This means that we can ignore all the lower-order terms, and simply consider the behavior of the highest-order terms.



In this case, the highest-order term in the numerator is 2x4, and the highest-order term in the denominator is x3. 



As x approaches infinity, 2x4 grows much faster than x3, and so the function f(x) approaches zero.



#



### These processes above  demonstrates how limits help us understand the behavior of functions near points that might not be defined, by finding equivalent expressions that are easier to evaluate.



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###### 
 Copyright 2024 Quantum Software Development. Code released under the [MIT License license.](https://github.com/Quantum-Software-Development/Math/blob/3bf8270ca09d3848f2bf22f9ac89368e52a2fb66/LICENSE)