Find limit of 2x+1-cosx/3x as x approaches 0?
To find the limit of the given expression as x approaches 0, let's analyze the different terms:
1. The term 2x approaches 0 as x approaches 0.
2. The term 1 approaches 1 as x approaches 0.
3. The term cos(x) approaches 1 as x approaches 0.
4. The term 3x approaches 0 as x approaches 0.
Therefore, the expression simplifies to:
(2x + 1 - cos(x)) / 3x
= (0 + 1 - 1) / 0
= 0 / 0
We have an indeterminate form of 0/0. This means we cannot directly substitute x = 0 into the expression to find the limit.
To proceed further, we need to apply L'Hôpital's Rule. This rule allows us to differentiate the numerator and the denominator separately and then take the limit again.
Let's differentiate the numerator and denominator:
Numerator:
d/dx (2x + 1 - cos(x))
= 2 - sin(x)
Denominator:
d/dx (3x)
= 3
Now, let's find the limit of the differentiated expression:
lim x→0 (2 - sin(x))/3
Substituting x = 0 into the differentiated expression:
(2 - sin(0))/3
= 2/3
So, the limit of (2x + 1 - cos(x))/3x as x approaches 0 is 2/3.
I will assume you meant:
lim (2x + 1 - cosx)/(3x)
then ...
= lim ( (2x)/(3x) + 1/(3x) - (1/3)(cosx/x) ) as x --> 0
= lim ( 2/3 + 0 - (1/3)(0) )
= 2/3