Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it.

lim x→0 (e^x − e^−x − 2x)/(x − sin(x))

you do have a 0/0 form

Using Hospital's rule
then lim= (e^x+e^-x-2)/(x+cosx)

Lim= 0/1=0

using L'Hospital's rule we have

lim x→0 (e^x − e^−x − 2x)/(x − sin(x))
lim x→0 (e^x + e^−x − 2)/(1 − cosx)
still 0/0, so do it again:
lim x→0 (e^x - e^−x)/sinx
and again:
lim x→0 (e^x + e^−x)/cosx
→ (1+1)/1 = 2

To find the limit of the given expression, we can apply l'Hospital's Rule since it is in the form of "0/0".

We will take the derivative of the numerator and denominator, then evaluate the limit of the resulting expression.

Taking the derivative of the numerator:
d/dx (e^x − e^(-x) − 2x) = e^x + e^(-x) - 2

Taking the derivative of the denominator:
d/dx (x − sin(x)) = 1 − cos(x)

Now, we can evaluate the limit of the expression:
lim x→0 (e^x + e^(-x) - 2) / (1 − cos(x))

Substituting x = 0 into the expression, we get:
(e^0 + e^(0) - 2) / (1 − cos(0))
(1 + 1 - 2) / (1 − 1)
0/0

Since l'Hospital's Rule can still be applied, we will take the derivatives again.

Taking the derivative of the numerator:
d/dx (e^x + e^(-x) - 2) = e^x - e^(-x)

Taking the derivative of the denominator:
d/dx (1 − cos(x)) = sin(x)

Now, we can evaluate the limit once again:
lim x→0 (e^x - e^(-x)) / sin(x)

Substituting x = 0 into the expression, we get:
(e^0 - e^(0)) / sin(0)
(1 - 1) / 0
0/0

Since we still have "0/0", we can once again apply l'Hospital's Rule and take the derivatives:

Taking the derivative of the numerator:
d/dx (e^x - e^(-x)) = e^x + e^(-x)

Taking the derivative of the denominator:
d/dx (sin(x)) = cos(x)

Now, we can evaluate the limit:
lim x→0 (e^x + e^(-x)) / cos(x)

Substituting x = 0 into the expression, we get:
(e^0 + e^(0)) / cos(0)
(1 + 1) / 1
2 / 1
2

Therefore, the limit of the given expression as x approaches 0 is 2.

To find the limit of the given expression as x approaches 0, we can use l'Hospital's Rule if applicable.

L'Hospital's Rule states that if we have a limit of the form 0/0 or ∞/∞, and if the derivatives of the numerator and denominator both exist or are infinite, then we can take the derivative of the numerator and denominator and evaluate the limit of the resulting expression. We can repeat this process until we obtain a limit that is not the form 0/0 or ∞/∞.

Let's apply l'Hospital's Rule to the given limit:

lim x→0 (e^x − e^−x − 2x)/(x − sin(x))

We can start by taking the derivatives of the numerator and denominator:

Numerator:
d/dx (e^x − e^−x − 2x) = e^x + e^(-x) - 2

Denominator:
d/dx (x − sin(x)) = 1 - cos(x)

Now, let's evaluate the limit of the derivatives:

lim x→0 (e^x + e^(-x) - 2)/(1 - cos(x))

Since taking the derivatives once did not remove the indeterminate form, we can apply l'Hospital's Rule again. Taking the derivatives of the numerator and denominator once more, we get:

Numerator:
d/dx (e^x + e^(-x) - 2) = e^x - e^(-x)

Denominator:
d/dx (1 - cos(x)) = sin(x)

Now, let's evaluate the limit of the second set of derivatives:

lim x→0 (e^x - e^(-x))/(sin(x))

Now we can try evaluating the limit without l'Hospital's Rule.
Observe that as x approaches 0, we have e^x → 1 and e^(-x) → 1. Also, sin(x) → 0.

Substituting these values, we get:

lim x→0 (1 - 1)/(0) = 0/0

Since we still have an indeterminate form, we can apply l'Hospital's Rule again. Taking the derivatives of the numerator and denominator:

Numerator:
d/dx (e^x - e^(-x)) = e^x + e^(-x)

Denominator:
d/dx (sin(x)) = cos(x)

Now, let's evaluate the limit of the third set of derivatives:

lim x→0 (e^x + e^(-x))/(cos(x))

Again, let's try evaluating the limit without l'Hospital's Rule.
As before, as x approaches 0, we have e^x → 1 and e^(-x) → 1. Additionally, cos(x) → 1.

Substituting these values, we get:

lim x→0 (1 + 1)/(1) = 2/1 = 2

Therefore, the limit of the given expression as x approaches 0 is 2.