MHB Find the limit as x approaches negative infinity....

[rayne1]

lim
x-> -(infinity) = x + sqrt(x^2 + 2x)

I know that you're supposed to multiply and divide it by it's conjugate and that the answer is -1. But I don't understand how the denominator x - sqrt(x^2 + 2x) = x + x*sqrt[1+(2/x^2)] = x[1+sqrt(1+(2/x^2)].

 

[I like Serena]

lim
x-> -(infinity) = x + sqrt(x^2 + 2x)

I know that you're supposed to multiply and divide it by it's conjugate and that the answer is -1. But I don't understand how the denominator x - sqrt(x^2 + 2x) = x + x*sqrt[1+(2/x^2)] = x[1+sqrt(1+(2/x^2)].

Hi rayne,

That doesn't look quite correct.

Can it be that the denominator should be:
$$x - \sqrt{x^2 + 2x} = x - x\sqrt{1+\frac 2x} = x\left(1-\sqrt{1+\frac 2x}\right)$$
(Wondering)

Edit: this is not right. See below.

 

[rayne1]

Hi rayne,

That doesn't look quite correct.

Can it be that the denominator should be:
$$x - \sqrt{x^2 + 2x} = x - x\sqrt{1+\frac 2x} = x\left(1-\sqrt{1+\frac 2x}\right)$$
(Wondering)

Okay then, my mistake. But I still don't understand how the 1st denominator changes to the other ones.

 

[I like Serena]

Okay then, my mistake. But I still don't understand how the 1st denominator changes to the other ones.

Let me split it up into smaller steps... and oops (Blush), I have just noticed I've made a mistake earlier:

$$x - \sqrt{x^2 + 2x} $$
$$= x - \sqrt{x^2\big(1+\frac 2x\big)} \tag 1$$
$$= x - \sqrt{x^2}\cdot\sqrt{1+\frac 2x} \tag 2$$
$$= x - |x|\cdot\sqrt{1+\frac 2x} \tag 3$$
$$= x + x\cdot\sqrt{1+\frac 2x} \tag 4$$
$$= x\left(1+\sqrt{1+\frac 2x}\right) \tag 5$$

Which step is it that you do not understand?

 

[rayne1]

Let me split it up into smaller steps... and oops (Blush), I have just noticed I've made a mistake earlier:

$$x - \sqrt{x^2 + 2x} $$
$$= x - \sqrt{x^2\big(1+\frac 2x\big)} \tag 1$$
$$= x - \sqrt{x^2}\cdot\sqrt{1+\frac 2x} \tag 2$$
$$= x - |x|\cdot\sqrt{1+\frac 2x} \tag 3$$
$$= x + x\cdot\sqrt{1+\frac 2x} \tag 4$$
$$= x\left(1+\sqrt{1+\frac 2x}\right) \tag 5$$

Which step is it that you do not understand?

Oh, I see. Now, why can't we leave the denominator as $$x - \sqrt{x^2 + 2x} $$ and divide each number by x^2 to get -1?

 

[I like Serena]

Oh, I see. Now, why can't we leave the denominator as $$x - \sqrt{x^2 + 2x} $$?

That's because we want to find the limit.

What do you have for the numerator? (Wondering)
That should help to make it clear.

 

[rayne1]

That's because we want to find the limit.

What do you have for the numerator? (Wondering)
That should help to make it clear.

I have it as -2x. Are we allowed to divide the numerator by x and the denominator by x^2?

 

[I like Serena]

I have it as -2x. Are we allowed to divide the numerator by x and the denominator by x^2?

No... but you can divide both by $x$.
What do you get if you do?

 

[rayne1]

No... but you can divide both by $x$.
What do you get if you do?

-2/1-sqrt(x+2)?

 

[I like Serena]

-2/1-sqrt(x+2)?

That should be:
$$\frac{-2x}{x(1+\sqrt{1+\frac 2x})}
=\frac{-2}{1+\sqrt{1+\frac 2x}}
$$
Like this only one $x$ is left, so we can finally see what happens if $x$ tends to $-\infty$.

 

[rayne1]

That should be:
$$\frac{-2x}{x(1+\sqrt{1+\frac 2x})}
=\frac{-2}{1+\sqrt{1+\frac 2x}}
$$
Like this only one $x$ is left, so we can finally see what happens if $x$ tends to $-\infty$.

Oh oops, I was a bit unclear. I thought we were talking about the first denominator (x - sqrt(x^2+2x).

 

[I like Serena]

Oh oops, I was a bit unclear. I thought we were talking about the first denominator (x - sqrt(x^2+2x).

That's the same denominator!
The whole point of rewriting the denominator, extracting $x$, is so that $x$ can be canceled against the $x$ in the numerator.

 

[Prove It]

lim
x-> -(infinity) = x + sqrt(x^2 + 2x)

I know that you're supposed to multiply and divide it by it's conjugate and that the answer is -1. But I don't understand how the denominator x - sqrt(x^2 + 2x) = x + x*sqrt[1+(2/x^2)] = x[1+sqrt(1+(2/x^2)].

$\displaystyle \begin{align*} x + \sqrt{ x^2 + 2x } &= x + \sqrt{ x^2 + 2x + 1 - 1 } \\ &= x + \sqrt{ \left( x + 1 \right) ^2 - 1 } \\ &= \ln{ \left( \mathrm{e}^{ x + \sqrt{ \left( x + 1 \right) ^2 - 1 } } \right) } \\ &= \ln{ \left( \mathrm{e}^{x}\,\mathrm{e}^{ \sqrt{ \left( x + 1 \right) ^2 - 1 } } \right) } \\ &= \ln{ \left( \frac{\mathrm{e}^{ \sqrt{ \left( x + 1 \right) ^2 - 1 } }}{\mathrm{e}^{ -x }} \right) } \end{align*}$

This is now an $\displaystyle \begin{align*} \frac{\infty}{\infty} \end{align*}$ indeterminate form, so you can use L'Hospital's Rule.