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

In summary: This is now an $\displaystyle \begin{align*} \frac{\infty}{\infty} \end{align*}$ indeterminate form, so you can use L'Hospital's Rule.
  • #1
rayne1
32
0
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)].
 
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  • #2
rayne said:
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.
 
  • #3
I like Serena said:
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.
 
  • #4
rayne said:
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:

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

Which step is it that you do not understand?
 
  • #5
I like Serena said:
Let me split it up into smaller steps... and oops (Blush), I have just noticed I've made a mistake earlier:

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

Which step is it that you do not understand?

Oh, I see. Now, why can't we leave the denominator as \(\displaystyle x - \sqrt{x^2 + 2x} \) and divide each number by x^2 to get -1?
 
  • #6
rayne said:
Oh, I see. Now, why can't we leave the denominator as \(\displaystyle x - \sqrt{x^2 + 2x} \)?

That's because we want to find the limit.

What do you have for the numerator? (Wondering)
That should help to make it clear.
 
  • #7
I like Serena said:
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?
 
  • #8
rayne said:
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?
 
  • #9
I like Serena said:
No... but you can divide both by $x$.
What do you get if you do?

-2/1-sqrt(x+2)?
 
  • #10
rayne said:
-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 $-\infty$.
 
  • #11
I like Serena said:
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).
 
  • #12
rayne said:
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.
 
  • #13
rayne said:
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.
 

FAQ: Find the limit as x approaches negative infinity....

What does "limit as x approaches negative infinity" mean?

The limit as x approaches negative infinity refers to the value that a function or sequence approaches as the input (x) gets closer and closer to negative infinity. In other words, it is the value that the function or sequence gets closer and closer to, but may not necessarily reach, as x gets smaller and smaller towards negative infinity.

How do you find the limit as x approaches negative infinity?

To find the limit as x approaches negative infinity, you can use the following steps:

  • Plug in a very large negative number for x in the given function or sequence.
  • If the resulting value approaches a specific number, that number is the limit as x approaches negative infinity.
  • If the resulting value approaches positive or negative infinity, then the limit does not exist.
  • If the resulting value fluctuates or does not approach a specific number, then you may need to use other methods such as L'Hopital's rule or graphing the function to determine the limit.

What does it mean if the limit as x approaches negative infinity is infinity?

If the limit as x approaches negative infinity is infinity, it means that the function or sequence grows without bound as x gets closer and closer to negative infinity. In other words, the function or sequence will continue to increase without a limit as x approaches negative infinity.

Can the limit as x approaches negative infinity be negative infinity?

Yes, the limit as x approaches negative infinity can be negative infinity. This means that the function or sequence decreases without bound as x gets closer and closer to negative infinity. In other words, the function or sequence will continue to decrease without a limit as x approaches negative infinity.

Is the limit as x approaches negative infinity always the same as the limit as x approaches positive infinity?

No, the limit as x approaches negative infinity and the limit as x approaches positive infinity can be different. This is because the behavior of a function or sequence can vary as x approaches different types of infinity. For example, a function may approach a different value as x approaches positive infinity compared to when x approaches negative infinity.

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