Limits, Sequences, Series: Learning Objectives 1, 2

Apply the Sum/Difference, Constant Function, and Power Laws for limits. Since the function is a polynomial, you can also use the Direct Substitution Property.

$39$

We start with the given limit: $$ \lim_{x \to 5} (2x^2 - 3x + 4) $$ Using the Limit Laws, we can break down the evaluation step by step:

1. Sum/Difference Law: The limit of a sum is the sum of the limits. $$ = \lim_{x \to 5} (2x^2) - \lim_{x \to 5} (3x) + \lim_{x \to 5} (4) $$
2. Constant Multiple Law: The limit of a constant times a function is the constant times the limit of the function. $$ = 2 \lim_{x \to 5} (x^2) - 3 \lim_{x \to 5} (x) + \lim_{x \to 5} (4) $$
3. Power Law and Constant Law: We can now substitute the value of $x$. $$ = 2(5^2) - 3(5) + 4 $$
4. Arithmetic: $$ = 2(25) - 15 + 4 = 50 - 15 + 4 = 39 $$

Since the function $f(x) = 2x^2-3x+4$ is a polynomial, we can also find the limit by direct substitution: $f(5) = 2(5)^2 - 3(5) + 4 = 39$.

This is a rational function. Check if direct substitution makes the denominator zero. If not, you can use the Quotient Law for limits and substitute the value directly.

$-\frac{1}{11}$

The function is a rational function. We can use the Direct Substitution Property if the denominator is not zero at the limit point.

1. Check the denominator: $$ \lim_{x \to -2} (5-3x) = 5 - 3(-2) = 5 + 6 = 11 $$ Since the denominator's limit is not 0, we can apply the Quotient Law.

2. Apply the Quotient Law and Direct Substitution: $$ \lim_{x \to -2} \frac{x^3+2x^2-1}{5-3x} = \frac{\lim_{x \to -2} (x^3+2x^2-1)}{\lim_{x \to -2} (5-3x)} $$ $$ = \frac{(-2)^3 + 2(-2)^2 - 1}{5 - 3(-2)} = \frac{-8 + 2(4) - 1}{11} = \frac{-8 + 8 - 1}{11} = -\frac{1}{11} $$

Direct substitution yields the indeterminate form $\frac{0}{0}$. Try to simplify the expression by factoring the numerator before taking the limit.

$2$

1. Attempt Direct Substitution: $$ \frac{1^2 - 1}{1 - 1} = \frac{0}{0} $$ This is an indeterminate form, which means we need to manipulate the expression algebraically.

2. Factor the Numerator: The numerator is a difference of squares, $a^2-b^2 = (a-b)(a+b)$. $$ \lim_{x \to 1} \frac{x^2-1}{x-1} = \lim_{x \to 1} \frac{(x-1)(x+1)}{x-1} $$ 3. Simplify: Since we are taking the limit as $x \to 1$, $x$ is not equal to 1, so we can safely cancel the $(x-1)$ terms. $$ = \lim_{x \to 1} (x+1) $$ 4. Evaluate the Limit: Now we can use direct substitution on the simplified expression. $$ = 1 + 1 = 2 $$

For a limit to exist, the left-hand limit and the right-hand limit must exist and be equal. Evaluate the limit as $x$ approaches 0 from the right ($x \to 0^+$) and from the left ($x \to 0^-$) separately.

The limit does not exist because the left- and right-hand limits are different.

To prove the limit does not exist, we must show that the left-hand limit is not equal to the right-hand limit.

1. Evaluate the right-hand limit ($x \to 0^+$):
When $x$ approaches 0 from the right, $x$ is positive, so $|x|=x$. $$ \lim_{x \to 0^+} \frac{|x|}{x} = \lim_{x \to 0^+} \frac{x}{x} = \lim_{x \to 0^+} 1 = 1 $$ 2. Evaluate the left-hand limit ($x \to 0^-$):
When $x$ approaches 0 from the left, $x$ is negative, so $|x|=-x$. $$ \lim_{x \to 0^-} \frac{|x|}{x} = \lim_{x \to 0^-} \frac{-x}{x} = \lim_{x \to 0^-} -1 = -1 $$ 3. Compare the limits:
Since $\lim_{x \to 0^+} \frac{|x|}{x} = 1$ and $\lim_{x \to 0^-} \frac{|x|}{x} = -1$, the one-sided limits are not equal.

Therefore, the overall limit $\lim_{x \to 0} \frac{|x|}{x}$ does not exist.

Think about the behavior of the denominator, $x^r$, as $x$ becomes an infinitely large positive number. What happens to the value of a fraction when its denominator grows without bound and the numerator is constant?

0

We are evaluating the limit of $f(x) = \frac{1}{x^r}$ as $x$ approaches infinity, given that $r$ is a positive rational number.

As $x \to \infty$, the term $x$ becomes arbitrarily large. Since $r > 0$, the term $x^r$ will also grow without bound. For example, if $r=2$, $x^2$ grows. If $r=1/2$, $\sqrt{x}$ also grows.

So, we have a fraction where the numerator is a constant (1) and the denominator is approaching infinity: $$ \lim_{x \to \infty} \frac{1}{x^r} \to \frac{1}{\infty} $$ As the denominator of a fraction grows infinitely large, the value of the fraction approaches 0.

Therefore, $\lim_{x \to \infty} \frac{1}{x^r} = 0$.

For limits of rational functions at infinity, a standard technique is to divide both the numerator and the denominator by the highest power of x found in the denominator.

$\frac{3}{5}$

1. Identify the highest power of x in the denominator: The denominator is $5x^2+4x+1$, so the highest power is $x^2$.

2. Divide every term by $x^2$: $$ \lim_{x \to \infty} \frac{\frac{3x^2}{x^2}-\frac{x}{x^2}-\frac{2}{x^2}}{\frac{5x^2}{x^2}+\frac{4x}{x^2}+\frac{1}{x^2}} = \lim_{x \to \infty} \frac{3-\frac{1}{x}-\frac{2}{x^2}}{5+\frac{4}{x}+\frac{1}{x^2}} $$ 3. Evaluate the limit of the new expression: We use the fact that $\lim_{x \to \infty} \frac{c}{x^r} = 0$ for any constant $c$ and $r>0$. $$ \frac{\lim_{x \to \infty} 3 - \lim_{x \to \infty} \frac{1}{x} - \lim_{x \to \infty} \frac{2}{x^2}}{\lim_{x \to \infty} 5 + \lim_{x \to \infty} \frac{4}{x} + \lim_{x \to \infty} \frac{1}{x^2}} = \frac{3 - 0 - 0}{5 + 0 + 0} = \frac{3}{5} $$ An alternative shortcut for rational functions is to compare the degrees of the numerator and denominator. Since they are equal (degree 2), the limit is the ratio of the leading coefficients: $\frac{3}{5}$.

This limit has the indeterminate form $\infty - \infty$. To resolve this, factor out the highest power of $x$ from the expression and analyze the resulting factors.

$\infty$

The limit $\lim_{x \to \infty} (x^2-x)$ is an indeterminate form because it involves subtracting one infinite quantity from another.

1. Factor out the highest power of x: In this case, we factor out $x^2$. $$ \lim_{x \to \infty} (x^2-x) = \lim_{x \to \infty} x^2 \left(1 - \frac{x}{x^2}\right) = \lim_{x \to \infty} x^2 \left(1 - \frac{1}{x}\right) $$ 2. Analyze the limit of each factor:

• $\lim_{x \to \infty} x^2 = \infty$
• $\lim_{x \to \infty} \left(1 - \frac{1}{x}\right) = 1 - 0 = 1$

3. Determine the overall limit:
We are multiplying a term that grows to infinity by a term that approaches 1. The result will grow to infinity. $$ (\infty) \times (1) \to \infty $$ Therefore, $\lim_{x \to \infty} (x^2-x) = \infty$.

The degree of the numerator is greater than the degree of the denominator. This suggests the limit will be infinite. To determine the sign, divide all terms by the highest power of $x$ in the denominator.

$-\infty$

1. Identify the highest power of x in the denominator: The denominator is $3-x$, so the highest power is $x$.

2. Divide every term by $x$: $$ \lim_{x \to \infty} \frac{\frac{x^2}{x} + \frac{x}{x}}{\frac{3}{x} - \frac{x}{x}} = \lim_{x \to \infty} \frac{x + 1}{\frac{3}{x} - 1} $$ 3. Analyze the numerator and denominator separately:

• Numerator: $\lim_{x \to \infty} (x+1) = \infty$
• Denominator: $\lim_{x \to \infty} (\frac{3}{x} - 1) = 0 - 1 = -1$

4. Determine the overall limit:
The expression becomes a numerator that grows to positive infinity divided by a denominator that approaches a negative constant (-1). $$ \frac{\infty}{-1} \to -\infty $$ Therefore, $\lim_{x \to \infty} \frac{x^2+x}{3-x} = -\infty$.

This is an indeterminate form of the type $\infty - \infty$. To simplify, multiply the expression by its conjugate over itself. The conjugate of $(\sqrt{A}-B)$ is $(\sqrt{A}+B)$.

$\frac{1}{6}$

1. Multiply by the Conjugate: The conjugate of $\sqrt{9x^2+x}-3x$ is $\sqrt{9x^2+x}+3x$. $$ \lim_{x \to \infty} \frac{(\sqrt{9x^2+x}-3x)(\sqrt{9x^2+x}+3x)}{\sqrt{9x^2+x}+3x} $$ 2. Simplify the Numerator: Use the difference of squares formula $(a-b)(a+b) = a^2-b^2$. $$ = \lim_{x \to \infty} \frac{(9x^2+x) - (3x)^2}{\sqrt{9x^2+x}+3x} = \lim_{x \to \infty} \frac{9x^2+x - 9x^2}{\sqrt{9x^2+x}+3x} = \lim_{x \to \infty} \frac{x}{\sqrt{9x^2+x}+3x} $$ 3. Divide by the Highest Power of x: The highest effective power is $x$. For the term in the square root, we divide by $x = \sqrt{x^2}$ (since $x>0$). $$ = \lim_{x \to \infty} \frac{\frac{x}{x}}{\frac{\sqrt{9x^2+x}}{\sqrt{x^2}} + \frac{3x}{x}} = \lim_{x \to \infty} \frac{1}{\sqrt{\frac{9x^2}{x^2} + \frac{x}{x^2}} + 3} = \lim_{x \to \infty} \frac{1}{\sqrt{9+\frac{1}{x}}+3} $$ 4. Evaluate the Limit: As $x \to \infty$, the term $\frac{1}{x} \to 0$. $$ = \frac{1}{\sqrt{9+0}+3} = \frac{1}{\sqrt{9}+3} = \frac{1}{3+3} = \frac{1}{6} $$

For a polynomial, the limit at infinity (or negative infinity) is determined by the behavior of the term with the highest degree.

$-\infty$

We are evaluating the limit of a polynomial as $x \to -\infty$. The end behavior of a polynomial is dominated by its term with the highest degree.

1. Identify the dominant term: In the polynomial $x^2+2x^7$, the term with the highest degree is $2x^7$.

2. Evaluate the limit of the dominant term: $$ \lim_{x \to -\infty} (x^2+2x^7) = \lim_{x \to -\infty} 2x^7 $$ 3. Analyze the behavior: As $x$ becomes a very large negative number, we consider the sign of the result.

• A negative number ($x$) raised to an odd power (7) results in a negative number ($x^7 \to -\infty$).
• Multiplying by a positive constant (2) does not change the sign.

Therefore, as $x \to -\infty$, the term $2x^7 \to -\infty$.