# Thread: Limit Problem with Infinity

1. ## Limit Problem with Infinity

$\dfrac{\lim}{x \to \infty}\dfrac{x^{2} - 4}{2 + x - 4x^{2}}$

Now divide everything by the highest index in the denominator

$\dfrac{\lim}{x \to \infty}\dfrac{x^{2}/ x^{2} - 4/x^{2}}{2/ x^{2}+ x/x^{2} - 4x^{2}/x^{2}}$

$\dfrac{\lim}{x \to \infty}\dfrac{1 -0}{0 + x^{-1} - 4}$ - See the $x^{-1}$ in the bottom?

$\dfrac{\lim}{x \to \infty}\dfrac{1 - 0 + x^{1}}{0 + - 4}$ - This becomes $x^{1}$ in the top. But how to evaluate to get $-\dfrac{1}{4}$

2. ## Re: Limit Problem with Infinity

Originally Posted by Jason76
$\dfrac{\lim}{x \to \infty}\dfrac{x^{2} - 4}{2 + x - 4x^{2}}$
$\dfrac{x^{2} - 4}{2 + x - 4x^{2}}=\dfrac{1 - 4x^{-2}}{2x^{-2} + x^{-1} - 4}$

3. ## Re: Limit Problem with Infinity

x to the power -1 is the same as 1/x, and as x tends to become infinitely large number, 1/x tends to become 0, hence it can simply be ignored in the denominator. Actually, that jump x to the power -1 made from denominator to numerator is a complete mathematical nonsense. Denominator should be factored and only then parts of such factored denominator could 'climb' over the fraction line (with its exponent changing sign) and multiply the numerator.

4. ## Re: Limit Problem with Infinity

Originally Posted by Jason76
$\dfrac{\lim}{x \to \infty}\dfrac{x^{2} - 4}{2 + x - 4x^{2}}$

Now divide everything by the highest index in the denominator

$\dfrac{\lim}{x \to \infty}\dfrac{x^{2}/ x^{2} - 4/x^{2}}{2/ x^{2}+ x/x^{2} - 4x^{2}/x^{2}}$

$\dfrac{\lim}{x \to \infty}\dfrac{1 -0}{0 + x^{-1} - 4}$ - See the $x^{-1}$ in the bottom?

$\dfrac{\lim}{x \to \infty}\dfrac{1 - 0 + x^{1}}{0 + - 4}$ - This becomes $x^{1}$ in the top. But how to evaluate to get $-\dfrac{1}{4}$
No, the " $x^{-1}$" does NOT become "x" in the top! You cannot move just one term from denominator to the numerator like that.

$\frac{1}{{1/5}- 4}$ is NOT equal to $\frac{5}{-4}$.

5. ## Re: Limit Problem with Infinity

Originally Posted by Jason76
$\dfrac{\lim}{x \to \infty}\dfrac{1 -0}{0 + x^{-1} - 4}$ - See the $x^{-1}$ in the bottom?

$\dfrac{\lim}{x \to \infty}\dfrac{1 - 0 + x^{1}}{0 + - 4}$ - This becomes $x^{1}$ in the top. But how to evaluate to get $-\dfrac{1}{4}$
What is your proof or what rule did you use to have this $\dfrac{a}{b^{-x}+c} = \dfrac{a+b^{x}}{c}$ as valid?

Also we don't write 0 + -4. You can't have 2 symbols for operations together. The correct way is 0+(-4).

6. ## Re: Limit Problem with Infinity

Originally Posted by Jason76
$\dfrac{\lim}{x \to \infty}\dfrac{x^{2} - 4}{2 + x - 4x^{2}}$

Now divide everything by the highest index in the denominator

$\dfrac{\lim}{x \to \infty}\dfrac{x^{2}/ x^{2} - 4/x^{2}}{2/ x^{2}+ x/x^{2} - 4x^{2}/x^{2}}$

$\dfrac{\lim}{x \to \infty}\dfrac{1 -0}{0 + x^{-1} - 4}$ - See the $x^{-1}$ in the bottom?

$\dfrac{\lim}{x \to \infty}\dfrac{1 - 0 + x^{1}}{0 + - 4}$ - This becomes $x^{1}$ in the top. But how to evaluate to get $-\dfrac{1}{4}$