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

lim x→0 (7x−sin 7x)/(7x−tan 7x)

To find the limit of the given expression, we can apply l'Hospital's Rule if it is appropriate. L'Hospital's Rule states that if we have an indeterminate form of the type 0/0 or ∞/∞, we can differentiate the numerator and the denominator separately and then take the limit again.

Let's differentiate the numerator and denominator:

The derivative of 7x is simply 7, and the derivative of sin(7x) is 7cos(7x) by the chain rule.
The derivative of 7x is simply 7, and the derivative of tan(7x) is 7sec^2(7x) by the chain rule.

Now we have the following expression:

lim x→0 (7 - 7cos(7x))/(7 - 7sec^2(7x))

Since the limit is still indeterminate (0/0), we can apply l'Hospital's Rule again.

Differentiating the numerator and denominator again, we obtain:

Derivative of 7 is 0, and the derivative of 7cos(7x) is -49sin(7x) by the chain rule.
Derivative of 7 is 0, and the derivative of 7sec^2(7x) is 98sec(7x)tan(7x) by the chain rule.

Now the expression becomes:

lim x→0 (-49sin(7x))/(98sec(7x)tan(7x))

At this point, we can simplify the expression. Notice that sec(7x)/tan(7x) is equal to cos(7x).

lim x→0 (-49sin(7x))/(98cos(7x))

Now we can take the limit directly by substituting x=0 into the expression:

(-49sin(0))/(98cos(0)) = 0/98 = 0

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