Let f(x)= 16x^5+13x^3+1/√(64x^10+10x^8+2x^4)

Find the limit as x→-∞ and x→∞

I know the answers are -2 and 2 now. But when I graph the expression, why does the calculator show the limits are -∞ and ∞?

Question

Let f(x)= 16x^5+13x^3+1/√(64x^10+10x^8+2x^4)

Find the limit as x→-∞ and x→∞

I know the answers are -2 and 2 now. But when I graph the expression, why does the calculator show the limits are -∞ and ∞?

experimentx · Answer

divide both sides by +inf ... the limit will be 2 for +inf

experimentx · Answer

should be -2 for -inf .. i guess .. but how to show it???

anonymous · Answer

+2 for $ +\infty$

anonymous · Answer

-2 for $-\infty $

anonymous · Answer

i havent done anything like this for ages but i think the following approach is ok:

multiply by $$\frac{x^{-5}}{x^{-5}}$$

to get

$$f(x) = \frac{16 + 13x^{-4} + x^{-5}}{\sqrt{64 + 10x^{-2} + 2x^{-6}}}$$

now let $$x \rightarrow \infty$$

f(x) tends to 2

anonymous · Answer

is that ok guys?

experimentx · Answer

yep ..

anonymous · Answer

You guys are right. How come when I graph the equation, f(x) increases toward infinity?

anonymous · Answer

@eigenschmeigen: That's amazing! :D

anonymous · Answer

what about negative, im not sure it works there

experimentx · Answer

give a try,,, factor our x from denominator.

anonymous · Answer

How come the graph doesn't back up -2 and 2?

anonymous · Answer

i have made an error, the numerator is wrong , should be 13 x^(-2)

experimentx · Answer

the dominant term is x power 5, which means it will be negative as x is negative

anonymous · Answer

why does my altered function not account for that?

anonymous · Answer

@eigenschmeigen: It doesn't matter, it's goint to be zero anyways :)

anonymous · Answer

yeah

anonymous · Answer

Read the second part of the question now

anonymous · Answer

still annoys me though. 

when x is negative after multiplying by (x^-5)/(x^-5) whats happening there?

experimentx · Answer

it doesn't come that way ... wanna keep trying??

anonymous · Answer

None of you can answer my real question...

anonymous · Answer

hang on ill wolfram it

anonymous · Answer

that will tell us whether its your calculator :)

anonymous · Answer

http://www.wolframalpha.com/input/?i=%2816x%5E5%2B13x%5E3%2B1%29%2F%E2%88%9A%2864x%5E10%2B10x%5E8%2B2x%5E4%29

anonymous · Answer

did you type it correctly?

experimentx · Answer

it's minus two .. definitely.

anonymous · Answer

of-course not.

anonymous · Answer

Ah, I needed parenthesis for the numerator too. Thank you.

anonymous · Answer

@FoolForMath , that last remark was not needed.

anonymous · Answer

i think the key is not to multiply by (x^-5)/(x^-5)

but by (x^-4)/(x^-4)

anonymous · Answer

that leaves x in the numerator

anonymous · Answer

and the denominator becoomes 8|x|

anonymous · Answer

which solves our negative problem

experimentx · Answer

@eigenschmeigen no ... i don't think so.

anonymous · Answer

@eigenschmeigen , no you factor out x^5 from the numberator and denominator.. you get 16+0+0 / √64+0+0

16/8 and 16/-8 are the answers.. there is no negative problem.

anonymous · Answer

how do we get the negative limit then?

anonymous · Answer

aahhh

experimentx · Answer

take out x^5 out of square root

anonymous · Answer

yes i see, from the top you factor out x^5 and |x^5| from the bottom

anonymous · Answer

ok

experimentx · Answer

yeah thats the trick ... tp add up ... it is safe to assume that the root taken out will be positive (since it is just out of square root) .... and sine function seems to be continuous.

anonymous · Answer

yeah that was silly of me xD