Find the limits in Exercises 17 22

Question

ScanSolve AI Tutor · Accepted Answer

here are the solutions for the remaining limits:

:

Substitute $x=-1$ into the expression.

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

Simplify the numerator and denominator.
$$ \frac{1 + 1 - 2}{-3} = \frac{0}{-3} $$

Perform the division.

$$
\frac{0}{-3} = 0
$$
The limit is $\boxed{\text{0}}$.

:

Direct substitution of $x=1$ results in $\frac{0}{0}$ , an indeterminate form.
Factor the numerator. $x^2 - 1 = (x - 1)(x + 1)$
Simplify the expression by canceling the common factor $(x - 1)$ .
$$ \lim_{x \to 1} \frac{(x - 1)(x + 1)}{x - 1} = \lim_{x \to 1} (x + 1) $$
Substitute $x=1$ into the simplified expression.
$$ 1 + 1 = 2 $$ The limit is $\boxed{\text{2}}$.
- :
Direct substitution of $x=-5$ results in $\frac{0}{0}$ , an indeterminate form.
Factor the numerator. $x^2 + 3x - 10 = (x + 5)(x - 2)$
Simplify the expression by canceling the common factor $(x + 5)$ .
$$ \lim_{x \to -5} \frac{(x + 5)(x - 2)}{x + 5} = \lim_{x \to -5} (x - 2) $$
Substitute $x=-5$ into the simplified expression.
$$ -5 - 2 = -7 $$ The limit is $\boxed{\text{-7}}$.
- :
Direct substitution of $x=2$ results in $\frac{0}{0}$ , an indeterminate form.
Factor the denominator. $x^2 + x - 6 = (x + 3)(x - 2)$
Simplify the expression by canceling the common factor $(x - 2)$ .
$$ \lim_{x \to 2} \frac{x - 2}{(x + 3)(x - 2)} = \lim_{x \to 2} \frac{1}{x + 3} $$

Substitute $x=2$ into the simplified expression.

$$
\frac{1}{2 + 3} = \frac{1}{5}
$$
The limit is $\boxed{\frac{1}{5}}$.

:

Direct substitution of $x=1$ results in $\frac{0}{0}$ , an indeterminate form.
Factor the numerator and denominator. $x^2 - 2x + 1 = (x - 1)^2$ $x^3 - 2x + 1 = (x - 1)(x^2 + x - 1)$

Simplify the expression by canceling the common factor $(x - 1)$ .

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

Substitute $x=1$ into the simplified expression.

$$
\frac{1 - 1}{1^2 + 1 - 1} = \frac{0}{1} = 0
$$
The limit is $\boxed{\text{0}}$.

:

Divide the numerator and denominator by the highest power of $x$ in the denominator, which is $x$ .

$$
\lim_{x \to \infty} \frac{2x + 3}{5x + 7} = \lim_{x \to \infty} \frac{\frac{2x}{x} + \frac{3}{x}}{\frac{5x}{x} + \frac{7}{x}} = \lim_{x \to \infty} \frac{2 + \frac{3}{x}}{5 + \frac{7}{x}}
$$

As $x \to \infty$ , terms of the form $\frac{c}{x^n}$ approach 0.
$$ \frac{2 + 0}{5 + 0} = \frac{2}{5} $$ The limit is $\boxed{\frac{2}{5}}$.
- :

Divide the numerator and denominator by the highest power of $x$ in the denominator, which is $x^2$ .

$$
\lim_{x \to \infty} \frac{2x^2 + 3}{5x^2 + 7} = \lim_{x \to \infty} \frac{\frac{2x^2}{x^2} + \frac{3}{x^2}}{\frac{5x^2}{x^2} + \frac{7}{x^2}} = \lim_{x \to \infty} \frac{2 + \frac{3}{x^2}}{5 + \frac{7}{x^2}}
$$

As $x \to \infty$ , terms of the form $\frac{c}{x^n}$ approach 0.
$$ \frac{2 + 0}{5 + 0} = \frac{2}{5} $$ The limit is $\boxed{\frac{2}{5}}$.
- :

Divide the numerator and denominator by the highest power of $x$ in the denominator, which is $x^3$ .

$$
\lim_{x \to \infty} \frac{x^2 - 4x + 8}{3x^3} = \lim_{x \to \infty} \frac{\frac{x^2}{x^3} - \frac{4x}{x^3} + \frac{8}{x^3}}{\frac{3x^3}{x^3}} = \lim_{x \to \infty} \frac{\frac{1}{x} - \frac{4}{x^2} + \frac{8}{x^3}}{3}
$$

As $x \to \infty$ , terms of the form $\frac{c}{x^n}$ approach 0.
$$ \frac{0 - 0 + 0}{3} = \frac{0}{3} = 0 $$ The limit is $\boxed{\text{0}}$.
- :
As $x \to \infty$ , the denominator $x^3 - 7x + 1$ approaches $\infty$ .
$$ \lim_{x \to \infty} \frac{1}{x^3 - 7x + 1} $$
A constant divided by an infinitely large number approaches 0.
$$ \frac{1}{\infty} = 0 $$ The limit is $\boxed{\text{0}}$.
- :

Divide the numerator and denominator by the highest power of $x$ in the denominator, which is $x$ .

$$
\lim_{x \to \infty} \frac{x^2 - 7x}{x + 1} = \lim_{x \to \infty} \frac{\frac{x^2}{x} - \frac{7x}{x}}{\frac{x}{x} + \frac{1}{x}} = \lim_{x \to \infty} \frac{x - 7}{1 + \frac{1}{x}}
$$

As $

Find the limits in Exercises 17 22

Step-by-step solution

Substitute $x=-1$ into the expression.

Simplify the numerator and denominator.

Perform the division.

Direct substitution of $x=1$ results in $\frac{0}{0}$ , an indeterminate form.

Simplify the expression by canceling the common factor $(x - 1)$ .

Substitute $x=1$ into the simplified expression.

Direct substitution of $x=-5$ results in $\frac{0}{0}$ , an indeterminate form.

Simplify the expression by canceling the common factor $(x + 5)$ .

Substitute $x=-5$ into the simplified expression.

Direct substitution of $x=2$ results in $\frac{0}{0}$ , an indeterminate form.

Simplify the expression by canceling the common factor $(x - 2)$ .

Substitute $x=2$ into the simplified expression.

Direct substitution of $x=1$ results in $\frac{0}{0}$ , an indeterminate form.

Simplify the expression by canceling the common factor $(x - 1)$ .

Substitute $x=1$ into the simplified expression.

Divide the numerator and denominator by the highest power of $x$ in the denominator, which is $x$ .

As $x \to \infty$ , terms of the form $\frac{c}{x^n}$ approach 0.

Divide the numerator and denominator by the highest power of $x$ in the denominator, which is $x^2$ .

As $x \to \infty$ , terms of the form $\frac{c}{x^n}$ approach 0.

Divide the numerator and denominator by the highest power of $x$ in the denominator, which is $x^3$ .

As $x \to \infty$ , terms of the form $\frac{c}{x^n}$ approach 0.

As $x \to \infty$ , the denominator $x^3 - 7x + 1$ approaches $\infty$ .

A constant divided by an infinitely large number approaches 0.

Divide the numerator and denominator by the highest power of $x$ in the denominator, which is $x$ .

As $

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Find the limits in Exercises 17 22

Step-by-step solution

Substitute x=−1x=-1x=−1 into the expression.

Simplify the numerator and denominator.

Perform the division.

Direct substitution of x=1x=1x=1 results in 00\frac{0}{0}00​, an indeterminate form.

Simplify the expression by canceling the common factor (x−1)(x - 1)(x−1).

Substitute x=1x=1x=1 into the simplified expression.

Direct substitution of x=−5x=-5x=−5 results in 00\frac{0}{0}00​, an indeterminate form.

Simplify the expression by canceling the common factor (x+5)(x + 5)(x+5).

Substitute x=−5x=-5x=−5 into the simplified expression.

Direct substitution of x=2x=2x=2 results in 00\frac{0}{0}00​, an indeterminate form.

Simplify the expression by canceling the common factor (x−2)(x - 2)(x−2).

Substitute x=2x=2x=2 into the simplified expression.

Direct substitution of x=1x=1x=1 results in 00\frac{0}{0}00​, an indeterminate form.

Simplify the expression by canceling the common factor (x−1)(x - 1)(x−1).

Substitute x=1x=1x=1 into the simplified expression.

Divide the numerator and denominator by the highest power of xxx in the denominator, which is xxx.

As x→∞x \to \inftyx→∞, terms of the form cxn\frac{c}{x^n}xnc​ approach 0.

Divide the numerator and denominator by the highest power of xxx in the denominator, which is x2x^2x2.

As x→∞x \to \inftyx→∞, terms of the form cxn\frac{c}{x^n}xnc​ approach 0.

Divide the numerator and denominator by the highest power of xxx in the denominator, which is x3x^3x3.

As x→∞x \to \inftyx→∞, terms of the form cxn\frac{c}{x^n}xnc​ approach 0.

As x→∞x \to \inftyx→∞, the denominator x3−7x+1x^3 - 7x + 1x3−7x+1 approaches ∞\infty∞.

A constant divided by an infinitely large number approaches 0.

Divide the numerator and denominator by the highest power of xxx in the denominator, which is xxx.

As $

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More Mathematics Questions

Substitute $x=-1$ into the expression.

Direct substitution of $x=1$ results in $\frac{0}{0}$ , an indeterminate form.

Simplify the expression by canceling the common factor $(x - 1)$ .

Substitute $x=1$ into the simplified expression.

Direct substitution of $x=-5$ results in $\frac{0}{0}$ , an indeterminate form.

Simplify the expression by canceling the common factor $(x + 5)$ .

Substitute $x=-5$ into the simplified expression.

Direct substitution of $x=2$ results in $\frac{0}{0}$ , an indeterminate form.

Simplify the expression by canceling the common factor $(x - 2)$ .

Substitute $x=2$ into the simplified expression.

Direct substitution of $x=1$ results in $\frac{0}{0}$ , an indeterminate form.

Simplify the expression by canceling the common factor $(x - 1)$ .

Substitute $x=1$ into the simplified expression.

Divide the numerator and denominator by the highest power of $x$ in the denominator, which is $x$ .

As $x \to \infty$ , terms of the form $\frac{c}{x^n}$ approach 0.

Divide the numerator and denominator by the highest power of $x$ in the denominator, which is $x^2$ .

As $x \to \infty$ , terms of the form $\frac{c}{x^n}$ approach 0.

Divide the numerator and denominator by the highest power of $x$ in the denominator, which is $x^3$ .

As $x \to \infty$ , terms of the form $\frac{c}{x^n}$ approach 0.

As $x \to \infty$ , the denominator $x^3 - 7x + 1$ approaches $\infty$ .

Divide the numerator and denominator by the highest power of $x$ in the denominator, which is $x$ .