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