Question

Evaluate Limx-->0 √(√x+x^3) . cos(π/x) uing sandwich theorem

Answer 1

Thanks Ishan ;)

Answer 2

btw Evaluate √ is only on (√x+x^3)

Answer 3

Oh okay.

Answer 4

\[\left(\sqrt{x} + x^3\right)\cos \frac{\pi}x\]?

Answer 5

No there is also a radical sign on (√x+x^3)

Answer 6

\[\sqrt{\left(\sqrt{x} + x^3\right)}\cos \frac{\pi}x\]

Answer 7

Hmm the Radical part is Zero at x-> 0

Answer 8

The range of the cosine function is from -1 to 1.

Answer 9

So. no matter what you put in as the domain cos or cosine function is limited by its range i.e [-1,1]. In my opinion the limit should approach Zero

Answer 10

http://calculus.nipissingu.ca/problems/limit_problems.html qustion 1 at the bottom of the page

Answer 11

Thats where I got this from

Answer 12

http://calculus.nipissingu.ca/problems/limit_problems.html

Answer 13

Hmm okay. but the answer should be Zero.

Answer 14

In the hint it says use sandwich theorem.The question has more to it ;)

Answer 15

However I think 0 is a reasonab;e answer.Thanks for your help anyway. Gve youa badge ;)

Answer 16

I don't know how to use Sandwhich theorem but the function cos is limited by its range [-1,1] and the radical part is Zero. So, the answer should be Zero.

Answer 17

I in grade 12 as well ;)

Answer 18

\[-1\le \cos t\le 1\]\[t = \frac{\pi}x\]\[-1\le \cos \frac{\pi}x\le1\]\[ \implies -\sqrt{\left(\sqrt{x} + x^3\right)}\le \sqrt{\left(\sqrt{x} + x^3\right)}\cos \frac{\pi}x\le \sqrt{\left(\sqrt{x} + x^3\right)}\]When \(x\to 0\).
\[0 \le \lim_{x\to0}\sqrt{\left(\sqrt{x} + x^3\right)}\cos\frac{\pi}x\le0\]\[\implies \lim_{x\to0}\sqrt{\left(\sqrt{x} + x^3\right)}\cos\frac{\pi}x\]
I like Squeeze Theorem or Sandwich Theorem (Just googled it)

Answer 19

The last eqation should be\[\implies \lim_{x\to0}\sqrt{\left(\sqrt{x} + x^3\right)}\cos\frac{\pi}x=0\]

Answer 20

equation*

Answer 21

I am prone to typos :(