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

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.

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

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.

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

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:

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

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

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.

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

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

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

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

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

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 $$\frac{1}{2}$$

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

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.

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

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.

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


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.

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

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.

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.

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:

$$x→∞lim x32x4−3x3+x+6 =0$$

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.