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1

Solved example of limits to infinity

$\lim_{x\to\infty}\left(\frac{2x^3-2x^2+x-3}{x^3+2x^2-x+1}\right)$

As a variable goes to infinity, the expression $2x^3-2x^2+x-3$ will behave the same way that it's largest power behaves

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

As a variable goes to infinity, the expression $x^3+2x^2-x+1$ will behave the same way that it's largest power behaves

$\frac{2x^3}{x^3}$

Plug in the value $\infty $ into the limit

$\frac{2\cdot \infty ^3}{\infty ^3}$

Infinity to the power of any positive number is equal to infinity, so $\infty ^3=\infty$

$\frac{2\infty }{\infty ^3}$

Any expression multiplied by infinity tends to infinity

$\frac{\infty }{\infty ^3}$

Infinity to the power of any positive number is equal to infinity, so $\infty ^3=\infty$

$\frac{\infty }{\infty }$
2

If we directly evaluate the limit $\lim_{x\to \infty }\left(\frac{2x^3-2x^2+x-3}{x^3+2x^2-x+1}\right)$ as $x$ tends to $\infty $, we can see that it gives us an indeterminate form

$\frac{\infty }{\infty }$
3

We can solve this limit by applying L'Hôpital's rule, which consists of calculating the derivative of both the numerator and the denominator separately

$\lim_{x\to \infty }\left(\frac{\frac{d}{dx}\left(2x^3-2x^2+x-3\right)}{\frac{d}{dx}\left(x^3+2x^2-x+1\right)}\right)$

Find the derivative of the numerator

$\frac{d}{dx}\left(2x^3-2x^2+x-3\right)$

The derivative of a sum of two functions is the sum of the derivatives of each function

$\frac{d}{dx}\left(2x^3\right)+\frac{d}{dx}\left(-2x^2+x-3\right)$

The derivative of a function multiplied by a constant ($2$) is equal to the constant times the derivative of the function

$2\frac{d}{dx}\left(x^3\right)+\frac{d}{dx}\left(-2x^2+x-3\right)$

The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$

$6x^{2}+\frac{d}{dx}\left(-2x^2+x-3\right)$

The derivative of a sum of two functions is the sum of the derivatives of each function

$6x^{2}+\frac{d}{dx}\left(-2x^2\right)+\frac{d}{dx}\left(x-3\right)$

The derivative of a function multiplied by a constant ($-2$) is equal to the constant times the derivative of the function

$6x^{2}-2\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(x-3\right)$

The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$

$6x^{2}-4x+\frac{d}{dx}\left(x-3\right)$

The derivative of a sum of two functions is the sum of the derivatives of each function

$6x^{2}-4x+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(-3\right)$

The derivative of the constant function ($-3$) is equal to zero

$6x^{2}-4x+\frac{d}{dx}\left(x\right)$

The derivative of the linear function is equal to $1$

$6x^{2}-4x+1$

Find the derivative of the denominator

$\frac{d}{dx}\left(x^3+2x^2-x+1\right)$

The derivative of a sum of two functions is the sum of the derivatives of each function

$\frac{d}{dx}\left(x^3\right)+\frac{d}{dx}\left(2x^2-x+1\right)$

The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$

$3x^{2}+\frac{d}{dx}\left(2x^2-x+1\right)$

The derivative of a sum of two functions is the sum of the derivatives of each function

$3x^{2}+\frac{d}{dx}\left(2x^2\right)+\frac{d}{dx}\left(-x+1\right)$

The derivative of a function multiplied by a constant ($2$) is equal to the constant times the derivative of the function

$3x^{2}+2\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(-x+1\right)$

The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$

$3x^{2}+4x+\frac{d}{dx}\left(-x+1\right)$

The derivative of a sum of two functions is the sum of the derivatives of each function

$3x^{2}+4x+\frac{d}{dx}\left(-x\right)+\frac{d}{dx}\left(1\right)$

The derivative of the constant function ($1$) is equal to zero

$3x^{2}+4x+\frac{d}{dx}\left(-x\right)$

The derivative of the linear function times a constant, is equal to the constant

$3x^{2}+4x-1$
4

After deriving both the numerator and denominator, the limit results in

$\lim_{x\to\infty }\left(\frac{6x^{2}-4x+1}{3x^{2}+4x-1}\right)$

As a variable goes to infinity, the expression $6x^{2}-4x+1$ will behave the same way that it's largest power behaves

$\frac{6x^{2}}{3x^{2}+4x-1}$

As a variable goes to infinity, the expression $3x^{2}+4x-1$ will behave the same way that it's largest power behaves

$\frac{6x^{2}}{3x^{2}}$

Plug in the value $\infty $ into the limit

$\frac{6\cdot \infty ^{2}}{3\cdot \infty ^{2}}$

Infinity to the power of any positive number is equal to infinity, so $\infty ^{2}=\infty$

$\frac{6\infty }{3\cdot \infty ^{2}}$

Any expression multiplied by infinity tends to infinity

$\frac{\infty }{3\cdot \infty ^{2}}$

Infinity to the power of any positive number is equal to infinity, so $\infty ^{2}=\infty$

$\frac{\infty }{3\infty }$

Any expression multiplied by infinity tends to infinity

$\frac{\infty }{\infty }$
5

If we directly evaluate the limit $\lim_{x\to \infty }\left(\frac{6x^{2}-4x+1}{3x^{2}+4x-1}\right)$ as $x$ tends to $\infty $, we can see that it gives us an indeterminate form

$\frac{\infty }{\infty }$
6

We can solve this limit by applying L'Hôpital's rule, which consists of calculating the derivative of both the numerator and the denominator separately

$\lim_{x\to \infty }\left(\frac{\frac{d}{dx}\left(6x^{2}-4x+1\right)}{\frac{d}{dx}\left(3x^{2}+4x-1\right)}\right)$

Find the derivative of the numerator

$\frac{d}{dx}\left(6x^{2}-4x+1\right)$

The derivative of a sum of two functions is the sum of the derivatives of each function

$\frac{d}{dx}\left(6x^{2}\right)+\frac{d}{dx}\left(-4x+1\right)$

The derivative of a function multiplied by a constant ($6$) is equal to the constant times the derivative of the function

$6\frac{d}{dx}\left(x^{2}\right)+\frac{d}{dx}\left(-4x+1\right)$

The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$

$12x+\frac{d}{dx}\left(-4x+1\right)$

The derivative of a sum of two functions is the sum of the derivatives of each function

$12x+\frac{d}{dx}\left(-4x\right)+\frac{d}{dx}\left(1\right)$

The derivative of the constant function ($1$) is equal to zero

$12x+\frac{d}{dx}\left(-4x\right)$

The derivative of the linear function times a constant, is equal to the constant

$12x-4$

Find the derivative of the denominator

$\frac{d}{dx}\left(3x^{2}+4x-1\right)$

The derivative of a sum of two functions is the sum of the derivatives of each function

$\frac{d}{dx}\left(3x^{2}\right)+\frac{d}{dx}\left(4x-1\right)$

The derivative of a function multiplied by a constant ($3$) is equal to the constant times the derivative of the function

$3\frac{d}{dx}\left(x^{2}\right)+\frac{d}{dx}\left(4x-1\right)$

The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$

$6x+\frac{d}{dx}\left(4x-1\right)$

The derivative of a sum of two functions is the sum of the derivatives of each function

$6x+\frac{d}{dx}\left(4x\right)+\frac{d}{dx}\left(-1\right)$

The derivative of the constant function ($-1$) is equal to zero

$6x+\frac{d}{dx}\left(4x\right)$

The derivative of the linear function times a constant, is equal to the constant

$6x+4$
7

After deriving both the numerator and denominator, the limit results in

$\lim_{x\to\infty }\left(\frac{12x-4}{6x+4}\right)$
8

Factor the numerator by $2$

$\lim_{x\to\infty }\left(\frac{2\left(6x-2\right)}{6x+4}\right)$
9

Factor the denominator by $2$

$\lim_{x\to\infty }\left(\frac{2\left(6x-2\right)}{2\left(3x+2\right)}\right)$
10

Cancel the common factor $2$

$\lim_{x\to\infty }\left(\frac{6x-2}{3x+2}\right)$

Plug in the value $\infty $ into the limit

$\frac{6\infty -2}{3\infty +2}$

Any expression multiplied by infinity tends to infinity

$\frac{\infty -2}{3\infty +2}$

Infinity plus any algebraic expression is equal to infinity

$\frac{\infty }{3\infty +2}$

Any expression multiplied by infinity tends to infinity

$\frac{\infty }{\infty +2}$

Infinity plus any algebraic expression is equal to infinity

$\frac{\infty }{\infty }$
11

If we directly evaluate the limit $\lim_{x\to \infty }\left(\frac{6x-2}{3x+2}\right)$ as $x$ tends to $\infty $, we can see that it gives us an indeterminate form

$\frac{\infty }{\infty }$
12

We can solve this limit by applying L'Hôpital's rule, which consists of calculating the derivative of both the numerator and the denominator separately

$\lim_{x\to \infty }\left(\frac{\frac{d}{dx}\left(6x-2\right)}{\frac{d}{dx}\left(3x+2\right)}\right)$

Find the derivative of the numerator

$\frac{d}{dx}\left(6x-2\right)$

The derivative of a sum of two functions is the sum of the derivatives of each function

$\frac{d}{dx}\left(6x\right)+\frac{d}{dx}\left(-2\right)$

The derivative of the constant function ($-2$) is equal to zero

$\frac{d}{dx}\left(6x\right)$

The derivative of the linear function times a constant, is equal to the constant

$6$

Find the derivative of the denominator

$\frac{d}{dx}\left(3x+2\right)$

The derivative of a sum of two functions is the sum of the derivatives of each function

$\frac{d}{dx}\left(3x\right)+\frac{d}{dx}\left(2\right)$

The derivative of the constant function ($2$) is equal to zero

$\frac{d}{dx}\left(3x\right)$

The derivative of the linear function times a constant, is equal to the constant

$3$
13

After deriving both the numerator and denominator, the limit results in

$\lim_{x\to\infty }\left(\frac{6}{3}\right)$
14

Divide $6$ by $3$

$\lim_{x\to\infty }\left(2\right)$
15

The limit of a constant is just the constant

$2$

Final Answer

$2$

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