find limits?

Question

lim e^x2 -1/cosx-1
x->0

Answer 1

this is of the form 0/0

so use l'hospitals rule

d/dx(numerator) = 2xe^2x
d/dx(denominator) = - sin x

ratio = -2(x/sin x) e^(x^2) = - 2.1.1 = -2 as e^ x^2 = 1 x/sinx =1 as x-> 0

Answer 2

I like doing this, this is using L' Hospitals rule, where you are supposed to take the derivitive of the top and bottom seperately, this is how it goes,

When a limit is given, you have to plug x into the equation and if it comes out to infinity/ infinity or 0/0 then you use the rule.

So in this case, when you plug in 0, you get 0/0, you can use the rule

2x(e^x^2)/ -sinx,

Now, after you take the derivitive of the top and bottom, you plug in zero again and see what you get.

0/ -sin0
0/0, So since we got 0/0 again you have to do the rule again.

2x(e^x^2)/ -sinx,
2x(2x(e^x^2))/-cosx,

Now when you plug in 0, on the top you get 0, and on the bottom you get 1.

0/1 = 0

So the limit is equal to zero.

Answer 3

0/0 form
using L Hospital's rule
limit=2xe^x^2/-sinx
=4x^2e^x^2+2e^x^2/-cosx
=2/-1=-2

Answer 4

lim e^x2 -1/cosx-1=e^0-1/(cos -1)=1-1/.5403=-.8508
x->0

Answer 5

-2:

A = lim e^x2 -1/cosx-1
x->0
A = - lim e^x2 -1/2sin^2(x/2)
x->0
A = - lim [e^x2 -1/x^2].[4(x/2)^2 / 2sin^2(x/2)]
x->0
We have:
lim e^x2 -1/x^2 =1
x->0

and

lim (x/2)^2 / sin^2(x/2) = 1
x->0

So A = - (4/2) = -2

Answer 6

Use l'Hopital's rule, I think.
When f(0)/g(0) = 0/0 then
take both derivatives and
evaluate limx->0 f'(x)/g'(x)