# Thread: Limits with Squeezing principle

1. ## Limits with Squeezing principle

I have to find the limit x-->0 arctan x cos(1/x), and justify my method, by using the squezzing princple. Can somebody help also I have another question(not as important because out teacher gave to us to think over the weekend but I have no idead where to start)

Let
f(x) = x^3 + bx^2 + cx + d be a general cubic polynomial with the
coefficient in front of
x3 adjusted to be a 1.
(a) Explain why
f(x) > 0 when x > 0 and very large, and why f(x) < 0

when
x < 0 and very large.
Hint. Rewrite
f as f(x) = x^3(1+b=x+c=x2+d=x2) and then inspect
the sign of the part in brackets when
x is very large.
(b) Use the above information to show that every cubic must cut the
x-axis
in at least one place, i.e. prove every cubic has a real root.

(c) Does every degree
4 polynomial cut the x-axis? Explain your answer.

2. I'll give you a hint for (c) for starters.

Both ends of a quartic go off in the same direction: either both to $y = + \infty$ or to $y = -\infty$.

So there's a point where it's at its minimum (in the first case) or maximum (in the second).

Add a big number to the first type of quartic so as to move the minimum above the line y = 0. Subtract a big one from the second so as to move the maximum below the line y = 0.

You will probably need to translate the above into more formal and rigorous language.

3. Originally Posted by hellohollar
I have to find the limit x-->0 arctan x cos(1=x), and justify my method, by using the squezzing princple. Can somebody help also I have another question(not as important because out teacher gave to us to think over the weekend but I have no idead where to start)

Is the function Arctan x * cos (x-1) or soemthing else? If is this one then the limit is zero since cox (x-1) --> cos(-1) when x --> 0 and Arctan x --> 0 when x --> 0. Now use arithmetic of limits

Let f(x) = x^3 + bx^2 + cx + d be a general cubic polynomial with the
coefficient in front of x3 adjusted to be a 1.
(a) Explain why f(x) > 0 when x > 0 and very large, and why f(x) < 0

when
x < 0 and very large.

As x --> oo or x --> -oo the highest power of x "rules over" the other powers of x. Since x^3 --> oo when x --> oo and x^3 --> -oo when x --> -oo the whole thing does the same.

The idea of the hint is very good too, if what's written in the brackets is a huge typo of yours and it says what I think it should.

Hint. Rewrite f as f(x) = x^3(1+b=x+c=x2+d=x2) and then inspect
the sign of the part in brackets when x is very large.
(b) Use the above information to show that every cubic must cut the x-axis
in at least one place, i.e. prove every cubic has a real root.

If on one side the cubic tends to -oo and on the other one it tends to +oo at some point it must be zero. this is intuitively obvious, but it is trivial using the intermediate value theorem for continuous functions.

(c) Does every degree 4 polynomial cut the x-axis? Explain your answer.

Of course not: x^4 + 1 is always positive...

Tonio

4. ## Wrong equation

Umm no the equation is arctanx * (cos (1/x))

5. What is that in response to?

For your first question, since cosine always has a value between -1 and 1,
$-arctan(x)\le arctan(x)cos(1/x)\le arctan(x)$. Now what is the limit of arctan(x) as x goes to 0?

For the second, surely your "hint" did NOT say "Rewrite f as $f(x) = x^3(1+b=x+c=x2+d=x2)$". That makes no sense. I suspect that is was "Rewrite f as $f(x)= x^3(1+ b/x+ c/x^2+ d/x^3)$". Now what happens to those fractions if x if very large, positive or negative?

For the last, look at Tonio's suggestion: y= x^4+ 1.

6. Originally Posted by hellohollar
Umm no the equation is arctanx * (cos (1/x))

Never mind: cos(1/x) is bounded and Arctan x --> 0 when x --> 0 so the limit still is 0

Tonio