Solved example of limits to infinity
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
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
Plug in the value $\infty $ into the limit
Infinity to the power of any positive number is equal to infinity, so $\infty ^3=\infty$
Any expression multiplied by infinity tends to infinity
Infinity to the power of any positive number is equal to infinity, so $\infty ^3=\infty$
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 $, we can see that it gives us an indeterminate form
We can solve this limit by applying L'Hôpital's rule, which consists of calculating the derivative of both the numerator and the denominator separately
Find the derivative of the numerator
The derivative of a sum of two functions is the sum of the derivatives of each function
The derivative of the constant function ($-3$) is equal to zero
The derivative of the linear function is equal to $1$
The derivative of a function multiplied by a constant ($-2$) is equal to the constant times the derivative of the function
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
The derivative of a function multiplied by a constant ($2$) is equal to the constant times the derivative of the function
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
Find the derivative of the denominator
The derivative of a sum of two functions is the sum of the derivatives of each function
The derivative of the constant function ($1$) is equal to zero
The derivative of the linear function times a constant, is equal to the constant
The derivative of a function multiplied by a constant ($2$) is equal to the constant times the derivative of the function
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
After deriving both the numerator and denominator, the limit results in
As a variable goes to infinity, the expression $-4x+1+6x^{2}$ will behave the same way that it's largest power behaves
As a variable goes to infinity, the expression $4x-1+3x^{2}$ will behave the same way that it's largest power behaves
Plug in the value $\infty $ into the limit
Infinity to the power of any positive number is equal to infinity, so $\infty ^{2}=\infty$
Any expression multiplied by infinity tends to infinity
Infinity to the power of any positive number is equal to infinity, so $\infty ^{2}=\infty$
Any expression multiplied by infinity tends to infinity
If we directly evaluate the limit $\lim_{x\to \infty }\left(\frac{-4x+1+6x^{2}}{4x-1+3x^{2}}\right)$ as $x$ tends to $\infty $, we can see that it gives us an indeterminate form
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
Find the derivative of the numerator
The derivative of a sum of two functions is the sum of the derivatives of each function
The derivative of the constant function ($1$) is equal to zero
The derivative of the linear function times a constant, is equal to the constant
The derivative of a function multiplied by a constant ($6$) is equal to the constant times the derivative of the function
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
Find the derivative of the denominator
The derivative of a sum of two functions is the sum of the derivatives of each function
The derivative of the constant function ($-1$) is equal to zero
The derivative of the linear function times a constant, is equal to the constant
The derivative of a function multiplied by a constant ($3$) is equal to the constant times the derivative of the function
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
After deriving both the numerator and denominator, the limit results in
Factor the numerator by $2$
Factor the denominator by $2$
Cancel the fraction's common factor $2$
Plug in the value $\infty $ into the limit
Any expression multiplied by infinity tends to infinity
Infinity plus any algebraic expression is equal to infinity
Any expression multiplied by infinity tends to infinity
Infinity plus any algebraic expression is equal to infinity
If we directly evaluate the limit $\lim_{x\to \infty }\left(\frac{-2+6x}{2+3x}\right)$ as $x$ tends to $\infty $, we can see that it gives us an indeterminate form
We can solve this limit by applying L'Hôpital's rule, which consists of calculating the derivative of both the numerator and the denominator separately
Find the derivative of the numerator
The derivative of a sum of two functions is the sum of the derivatives of each function
The derivative of the constant function ($-2$) is equal to zero
The derivative of the linear function times a constant, is equal to the constant
Find the derivative of the denominator
The derivative of a sum of two functions is the sum of the derivatives of each function
The derivative of the constant function ($2$) is equal to zero
The derivative of the linear function times a constant, is equal to the constant
Divide $6$ by $3$
After deriving both the numerator and denominator, the limit results in
The limit of a constant is just the constant
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