Calcolatrice di Limiti secondo la regola di L'Hpital

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 Esempio

 Problemi risolti

 Problemi difficili

Here, we show you a step-by-step solved example of limits by l'hôpital's rule. This solution was automatically generated by our smart calculator:

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

Plug in the value $0$ into the limit

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

The cosine of $0$ equals $1$

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

Multiply $-1$ times $1$

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

Subtract the values $1$ and $-1$

$\lim_{x\to 0}\left(\frac{0}{0^2}\right)$

Calculate the power $0^2$

$\lim_{x\to 0}\left(\frac{0}{0}\right)$

If we directly evaluate the limit $\lim_{x\to 0}\left(\frac{1-\cos\left(x\right)}{x^2}\right)$ as $x$ tends to $0$, we can see that it gives us an indeterminate form

$\frac{0}{0}$

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

$\lim_{x\to 0}\left(\frac{\frac{d}{dx}\left(1-\cos\left(x\right)\right)}{\frac{d}{dx}\left(x^2\right)}\right)$

Find the derivative of the numerator

$\frac{d}{dx}\left(1-\cos\left(x\right)\right)$

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

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

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

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

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

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

The derivative of the cosine of a function is equal to minus the sine of the function times the derivative of the function, in other words, if $f(x) = \cos(x)$, then $f'(x) = -\sin(x)\cdot D_x(x)$

$1\sin\left(x\right)$

Any expression multiplied by $1$ is equal to itself

$\sin\left(x\right)$

Find the derivative of the denominator

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

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

$2x$

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

$\lim_{x\to 0}\left(\frac{\sin\left(x\right)}{2x}\right)$

Plug in the value $0$ into the limit

$\lim_{x\to 0}\left(\frac{\sin\left(0\right)}{2\cdot 0}\right)$

The sine of $0$ equals $0$

$\lim_{x\to 0}\left(\frac{0}{2\cdot 0}\right)$

Multiply $2$ times $0$

$\lim_{x\to 0}\left(\frac{0}{0}\right)$

If we directly evaluate the limit $\lim_{x\to 0}\left(\frac{\sin\left(x\right)}{2x}\right)$ as $x$ tends to $0$, we can see that it gives us an indeterminate form

$\frac{0}{0}$

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

$\lim_{x\to 0}\left(\frac{\frac{d}{dx}\left(\sin\left(x\right)\right)}{\frac{d}{dx}\left(2x\right)}\right)$

Find the derivative of the numerator

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

The derivative of the sine of a function is equal to the cosine of that function times the derivative of that function, in other words, if ${f(x) = \sin(x)}$, then ${f'(x) = \cos(x)\cdot D_x(x)}$

$\cos\left(x\right)$

Find the derivative of the denominator

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

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

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

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

$2$

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

$\lim_{x\to 0}\left(\frac{\cos\left(x\right)}{2}\right)$

Evaluate the limit $\lim_{x\to 0}\left(\frac{\cos\left(x\right)}{2}\right)$ by replacing all occurrences of $x$ by $0$

$\frac{\cos\left(0\right)}{2}$

The cosine of $0$ equals $1$

$\frac{1}{2}$

 Risposta finale al problema