(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

[tex] \lim_{t \to -\infty} \frac {\sqrt{t^{2}+2}}{4t+2} [/tex]

2. Relevant equations

3. The attempt at a solution

Later in the problem I will use:

[tex] \epsilon^{+} \;\;\;\;\;\;\;and\;\;\;\;\;\;\; \epsilon^{-} [/tex]

to represent positive and negative infinitesimals, respectively.

So:

[tex] \lim_{t \to -\infty} \frac {\sqrt{t^{2}+2}}{4t+2} [/tex]

[tex] \lim_{t \to -\infty} \frac {\sqrt{t^{2}(1+2t^{-2})}}{4t+2} [/tex]

[tex] \lim_{t \to -\infty} \frac {t\sqrt{1+2t^{-2}}}{4t+2} [/tex]

[tex] \lim_{t \to -\infty} \frac {t\sqrt{1+2t^{-2}}}{t(4+2t^{-1})} [/tex]

[tex] \lim_{t \to -\infty} \frac {\sqrt{1+2t^{-2}}}{4+2t^{-1}} [/tex]

So at this point the limit is basically saying this:

[tex] \frac {1 + \epsilon^{+}}{4+ \epsilon^{-}} [/tex]

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:

[tex] \lim_{t \to -\infty} \frac {\sqrt{t^{2}+2}}{4t+2} = \frac {1}{4} [/tex]

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?

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# Homework Help: Infinite Limit help

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