Quantum-Software-Development/Limits-Calculus-I

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

✍️ Limits - Calculus I - Resolution of Mathematics Exercises

AI Data Science - 1st Semester / 20024 - PUCSP University - Math Repository - Professor Eric Bacconi Gonçalves

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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