Help darivatives?

Question

lim [(e^x)*cos(x) -1 - x ] / [(x^2) + (x^3)]
x--->0


i got the derivative, but i cant substitute 0 into it, else it will be undefined(Im not chuck norris, i cant divide by 0 lol)

Answer 1

lim [ e^x cos(x) - 1 - x ] / [ x^2 + x^3 ]
x -> 0

This is of the indeterminate form [0/0]. Use L'Hospital's rule. This means using the product rule on e^x cos(x).

lim [ e^x cos(x) + e^x (-sin(x)) - 1 ] / [ 2x + 3x^2 ]
x -> 0

This simplifies as

lim [ e^x cos(x) - e^x sin(x) - 1 ] / [ 2x + 3x^2 ]
x -> 0

Factor out e^x from the first two terms.

lim [ e^x (cos(x) - sin(x)) - 1 ] / [2x + 3x^2 ]
x -> 0

This is of the form [0/0]. Use L'Hospital's rule again, and the product rule again.

lim [ e^x (cos(x) - sin(x)) + e^x (-sin(x) - cos(x)) ] / [2 + 6x]
x -> 0

Which we can simplify as

lim [ e^x cos(x) - e^x sin(x) - e^x sin(x) - e^x cos(x) ] / [2 + 6x]
x -> 0

Group like terms; we get a cancellation.

lim [ -2e^x sin(x) ] / [2 + 6x]
x -> 0

Now we can safely substitute x = 0 and evaluate the limit.

[ -2e^0 sin(0) ] / [2 + 6*0]
[-2(1)(0)] / [2 + 0]
0/2

Which is just

0

Answer 2

We know, from Taylor theorem, that, when x -> 0, then

e^x =~ 1 + x + x^2/2
cos(x) =~ 1- x^2/2 Therefore, the given expression, for x ->0, x <>0, is equivalent to

((1+x + x^2/2)(1 - x^2/2)) -1 - x)/(x^2 + x^3) = (1 - x^2/2 + x - x^3/2 +x^2/2 -x^4/4 - 1 -x)/(x^2 + x^3) = ( - x^3/2 - x^4/4)/(x^2 + x^3) = (-x/2 - x^2/4)/(1 + x). So, when x-> 0 the numerator goes to 0 and the and the denominator goes to 1, and our limit is 0/1 = 0.

You could as well use L'Hopital Rule and differentiate twice to get rid of a null denominator at x = 0

Answer 3

It is possible to differentiate the numerator and the denominator several times. Untill you get an expression which is defined when x = 0.
So, differentiating first time you get:
[(e^x)*cos(x)-(e^x)*sin(x)-1]
-------------------------------------
[2x+3(x^2)]

Applying differentiation once more:
- 2(e^x)*sin(x)/[2+6x]

And when x tends to 0 you get 0 as the limit value.