Limits by factoring Calculator

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Difficult Problems

1

Solved example of limits by factoring

$\lim_{x\to4}\left(\frac{x^2-16}{x^2+2x-24}\right)$

Plug in the value $4$ into the limit

$\frac{4^2-16}{4^2+2\cdot 4-24}$

Calculate the power $4^2$

$\frac{16-16}{4^2+2\cdot 4-24}$

Subtract the values $16$ and $-16$

$\frac{0}{4^2+2\cdot 4-24}$

Multiply $2$ times $4$

$\frac{0}{4^2+8-24}$

Subtract the values $8$ and $-24$

$\frac{0}{-16+4^2}$

Calculate the power $4^2$

$\frac{0}{-16+16}$

Subtract the values $16$ and $-16$

$\frac{0}{0}$
2

If we directly evaluate the limit $\lim_{x\to 4}\left(\frac{x^2-16}{x^2+2x-24}\right)$ as $x$ tends to $4$, we can see that it gives us an indeterminate form

$\frac{0}{0}$
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

$\lim_{x\to 4}\left(\frac{\frac{d}{dx}\left(x^2-16\right)}{\frac{d}{dx}\left(x^2+2x-24\right)}\right)$

Find the derivative of the numerator

$\frac{d}{dx}\left(x^2-16\right)$

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

$\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(-16\right)$

The derivative of the constant function ($-16$) is equal to zero

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

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

$2x$

Find the derivative of the denominator

$\frac{d}{dx}\left(x^2+2x-24\right)$

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

$\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(2x\right)+\frac{d}{dx}\left(-24\right)$

The derivative of the constant function ($-24$) is equal to zero

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

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

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

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

$2x+2$
4

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

$\lim_{x\to4}\left(\frac{2x}{2x+2}\right)$
5

Factor the denominator by $2$

$\lim_{x\to4}\left(\frac{2x}{2\left(x+1\right)}\right)$
6

Cancel the fraction's common factor $2$

$\lim_{x\to4}\left(\frac{x}{x+1}\right)$
7

Evaluate the limit $\lim_{x\to4}\left(\frac{x}{x+1}\right)$ by replacing all occurrences of $x$ by $4$

$\frac{4}{4+1}$

Add the values $4$ and $1$

$\frac{4}{5}$

Divide $4$ by $5$

$\frac{4}{5}$
8

Simplifying, we get

$\frac{4}{5}$

$\frac{4}{5}$