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  1. Rechenmaschinen
  2. Grenzen Der Unendlichkeit

Grenzen der Unendlichkeit Rechner

Mit unserem Grenzen der Unendlichkeit Schritt-für-Schritt-Rechner erhalten Sie detaillierte Lösungen für Ihre mathematischen Probleme. Üben Sie Ihre mathematischen Fähigkeiten und lernen Sie Schritt für Schritt mit unserem Mathe-Löser. Alle unsere Online-Rechner finden Sie hier.

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Beispiel

Gelöste Probleme

Schwierige Probleme

1

Here, we show you a step-by-step solved example of limits to infinity. This solution was automatically generated by our smart calculator:

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

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

$2x^3$

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

$2x^2$

Plug in the value $\infty $ into the limit

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

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

$\lim_{x\to\infty }\left(\frac{2\cdot \infty }{2\cdot \infty ^2}\right)$

Any expression multiplied by infinity tends to infinity, in other words: $\infty\cdot(\pm n)=\pm\infty$, if $n\neq0$

$\lim_{x\to\infty }\left(\frac{\infty }{2\cdot \infty ^2}\right)$

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

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

Any expression multiplied by infinity tends to infinity, in other words: $\infty\cdot(\pm n)=\pm\infty$, if $n\neq0$

$\lim_{x\to\infty }\left(\frac{\infty }{\infty }\right)$
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 or more functions is the sum of the derivatives of each function

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

The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function

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

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

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

Multiply $2$ times $3$

$6x^{2}-2\cdot 2x+1$

Multiply $-2$ times $2$

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

Find the derivative of the denominator

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

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

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

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

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

The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function

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

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

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

Multiply $2$ times $2$

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

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

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

After deriving both the numerator and denominator, and simplifying, 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

$6x^{2}$

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

$3x^{2}$

Plug in the value $\infty $ into the limit

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

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

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

Any expression multiplied by infinity tends to infinity, in other words: $\infty\cdot(\pm n)=\pm\infty$, if $n\neq0$

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

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

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

Any expression multiplied by infinity tends to infinity, in other words: $\infty\cdot(\pm n)=\pm\infty$, if $n\neq0$

$\lim_{x\to\infty }\left(\frac{\infty }{\infty }\right)$
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 or more functions is the sum of the derivatives of each function

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

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

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

The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function

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

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

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

Multiply $6$ times $2$

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

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

$12x-4$

Find the derivative of the denominator

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

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

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

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

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

The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function

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

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

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

Multiply $3$ times $2$

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

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

$6x+4$

Factor the numerator by $2$

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

Factor the denominator by $2$

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

Cancel the fraction's common factor $2$

$\lim_{x\to\infty }\left(\frac{6x-2}{3x+2}\right)$
7

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

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

Plug in the value $\infty $ into the limit

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

Any expression multiplied by infinity tends to infinity, in other words: $\infty\cdot(\pm n)=\pm\infty$, if $n\neq0$

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

Infinity plus any algebraic expression is equal to infinity

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

Any expression multiplied by infinity tends to infinity, in other words: $\infty\cdot(\pm n)=\pm\infty$, if $n\neq0$

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

Infinity plus any algebraic expression is equal to infinity

$\lim_{x\to\infty }\left(\frac{\infty }{\infty }\right)$
8

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

$\frac{\infty }{\infty }$
9

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 or more functions is the sum of the derivatives of each function

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

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

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

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

$6$

Find the derivative of the denominator

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

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

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

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

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

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

$3$

Divide $6$ by $3$

$\lim_{x\to\infty }\left(2\right)$
10

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

$\lim_{x\to\infty }\left(2\right)$
11

The limit of a constant is just the constant

$2$

Endgültige Antwort auf das Problem

$2$

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