Solve Infinite Limit: Find $\frac {1}{4}$

[Asphyxiated]

Homework Statement

$$\lim_{t \to -\infty} \frac {\sqrt{t^{2}+2}}{4t+2}$$

Homework Equations

The Attempt at a Solution

Later in the problem I will use:

$$\epsilon^{+} \;\;\;\;\;\;\;and\;\;\;\;\;\;\; \epsilon^{-}$$

to represent positive and negative infinitesimals, respectively.

So:

$$\lim_{t \to -\infty} \frac {\sqrt{t^{2}+2}}{4t+2}$$

$$\lim_{t \to -\infty} \frac {\sqrt{t^{2}(1+2t^{-2})}}{4t+2}$$

$$\lim_{t \to -\infty} \frac {t\sqrt{1+2t^{-2}}}{4t+2}$$

$$\lim_{t \to -\infty} \frac {t\sqrt{1+2t^{-2}}}{t(4+2t^{-1})}$$

$$\lim_{t \to -\infty} \frac {\sqrt{1+2t^{-2}}}{4+2t^{-1}}$$

So at this point the limit is basically saying this:

$$\frac {1 + \epsilon^{+}}{4+ \epsilon^{-}}$$

this is because the reciprocal of negative infinity squared is a positive infinitesimal and the reciprocal of negative infinity is a negative infinitesimal so the limit is:

$$\lim_{t \to -\infty} \frac {\sqrt{t^{2}+2}}{4t+2} = \frac {1}{4}$$

My book states it should be -1/4 but I do not see why. Your adding a positive infinitesimal to 1 and subtracting it from 4, but that doesn't make the numerator or denominator negative, so what I am I missing?

 

[physicsman2]

Delete post

 

[Dickfore]

$$\sqrt{t^{2}} = \left| t \right| = \left\{<br /> \begin{array}{lcc}<br /> t & , & t \ge 0 \\<br /> -t & , & t < 0<br /> \end{array} \right.$$

 

[physicsman2]

$$\sqrt{t^{2}} = \left| t \right| = \left\{<br /> \begin{array}{lcc}<br /> t & , & t \ge 0 \\<br /> -t & , & t < 0<br /> \end{array} \right.$$

That too. So the above is a better answer. Multiply by |t|/|t| and look at the situation, ie. where the limit is going off to, if you have a problem like the one you have with a square root, etc. in order to see if you need a +t or -t.

 

[Dickfore]

No! You only made a mistake in step 3 where you took $t^{2}$ from the square root. According to my formula, you should have taken something else. Everything else follows from there.

 

[physicsman2]

No! You only made a mistake in step 3 where you took $t^{2}$ from the square root. According to my formula, you should have taken something else. Everything else follows from there.

You're right, I feel stupid now for not realizing my mistake from the beginning. They should have left the t^2 in there and multiplied the expression by |t|/|t| to take the t^2 away in the numerator and leave a -t to be multiplied through in the denominator and apply the limit to get the answer.

 

[Asphyxiated]

Could you possibly explain what it is that I was suppose to take out? The (1+2t^{-2})? cause that would be the only part to take out of the radical and I don't see that getting me to the correct solution.

@physicsman2

I don't see why I would need to multiply t/t by -t/-t to get them to cancel just because it is -infinity, t/t = t*t^-1 = 1 inherently, why is the multiplication necessary?

 

[Asphyxiated]

sorry, didnt notice the last post physicsman2,

so should i not have even factored out the t^2 and instead multiplied by |t|/|t| which is effectively -t because the limit is - infinity? I still don't see how it is solved though...

 

[Tedjn]

You solved it, though. When t < 0, |t|/t = -1, which is where the negative appears.

 

[Asphyxiated]

oh, i am pretty sure i got it then

 

[physicsman2]

You solved it, though. When t < 0, |t|/t = -1, which is where the negative appears.

Pretty much this.

Sorry for the confusion earlier, I wasn't thinking all that well until I was corrected.
Remember that |t| also equals sqrt(t^2). |t| can also equal either t or -t the way Dickfore showed. Since you want to take out a t^2 and cancel to apply the limit, you change |t| to sqrt(t^2) to bring it into the radical and divide. You would change the |t| in the denominator to -t because you're approaching -infinity, just as Tedjn showed. Divide and apply the limit, and you get -1/4