Calc. I - Limits (x-->infinity)

[math-is-fun]

Hi guys. Quick beginners question. What is the limit of sqrt(9t^2 + 0.5t + 179)/(0.2t + 1500) as t approaches + infinity. This is pulled from my gf's calculus I text: ch. 1. And so, the only procedures we're currently aware of for finding limits as c--> +/- infinity is restricted to "reciprocal limit rules", other general means of algebraic simplification, and simply plugging in t-values, though we could very well be forgetting something. We can't seem to manipulate it into the book's clean-cut answer of 1500. And oddly enough, when we stick the function into online limit calculators it seems to simplify to 15. Perhaps the book just expects us to plug in larger and larger x values, but we'd really prefer an algebraic solution. Thanks for taking the time to read this. Any help is greatly appreciated!

-Matt -n- Sarah

 

[Deleted member 4993]

Hi guys. Quick beginners question. What is the limit of sqrt(9t^2 + 0.5t + 179)/(0.2t + 1500) as t approaches + infinity. This is pulled from my gf's calculus I text: ch. 1. And so, the only procedures we're currently aware of for finding limits as c--> +/- infinity is restricted to "reciprocal limit rules", other general means of algebraic simplification, and simply plugging in t-values, though we could very well be forgetting something. We can't seem to manipulate it into the book's clean-cut answer of 1500. And oddly enough, when we stick the function into online limit calculators it seems to simplify to 15. Perhaps the book just expects us to plug in larger and larger x values, but we'd really prefer an algebraic solution. Thanks for taking the time to read this. Any help is greatly appreciated!

-Matt -n- Sarah

The answer is 15 - unless you made a mistake in looking up the answer or copying the problem.

\(\displaystyle \dfrac{\sqrt{9t^2+0.5t+179}}{0.2t+1500}\)

divide the numerator and the denominator by 't'

\(\displaystyle = \ \dfrac{\sqrt{9 \ + \ \frac{0.5}{t} \ + \ \frac{179}{t^2}}}{0.2 \ + \ \frac{1500}{t}}\)

Now take the limit as t→ ∞

 

[fcabanski]

The answer is 15. The book answer, 1500, is not correct. You can verify it by graphing the function.

The answer is 15 - unless you made a mistake in looking up the answer or copying the problem.

\(\displaystyle \dfrac{\sqrt{9t^2+0.5t+179}}{0.2t+1500}\)

divide the numerator and the denominator by 't'

\(\displaystyle = \ \dfrac{\sqrt{9 \ + \ \frac{0.5}{t} \ + \ \frac{179}{t^2}}}{0.2 \ + \ \frac{1500}{t}}\)

Now take the limit as t→ ∞

 

[math-is-fun]

Thanks

The answer is 15 - unless you made a mistake in looking up the answer or copying the problem.

\(\displaystyle \dfrac{\sqrt{9t^2+0.5t+179}}{0.2t+1500}\)

divide the numerator and the denominator by 't'

\(\displaystyle = \ \dfrac{\sqrt{9 \ + \ \frac{0.5}{t} \ + \ \frac{179}{t^2}}}{0.2 \ + \ \frac{1500}{t}}\)

Now take the limit as t→ ∞

Thanks for the clear and concise answers. We weren't exactly sure as to how to divide t into the radical, but it all makes sense now. And yes--I actually omitted the units, which is "a thousand dollars per person": referring to per capita earnings of some given country (per capita earnings as a function of time). And so, that's where the book's answer of $1500 per person comes from. Sorry about leaving that part out but thanks again for the quick responses!!

 

[Quaid]

We weren't exactly sure as to how to divide t into the radical, but it all makes sense now.

Quick note and warning.

We can divide the radicand in the numerator by t^2 because we know that t is always a positive value.

Be careful, in cases where t approaches negative infinity.

:cool:

 

[Deleted member 4993]

Thanks for the clear and concise answers. We weren't exactly sure as to how to divide t into the radical, but it all makes sense now. And yes--I actually omitted the units, which is "a thousand dollars per person": referring to per capita earnings of some given country (per capita earnings as a function of time). And so, that's where the book's answer of $1500 per person comes from. Sorry about leaving that part out but thanks again for the quick responses!!

In that case the book's answer should have been:

$15000 per person - yes?