For this problem it's important to notice that e is raised to the power of negative x.  

It may be clearer to rewrite the function as:

As x grows large the function will approach zero.

Recall the definition of the derivative for a function f(x):

This limit returns the derivative of f(x). For the limit presented in the problem, f(x)=sinx and f(x+h)=sin(x+h), so it's really just asking us to take the derivative of sinx which is trivial.  

As the function approaches infinity, both the numerator and the denominator approach infinity.

This is an indeterminate form so we can apply L'Hospital's rule to take the limit.  After taking the derivative of the numerator and denominator we get the following limit:

Since the denominator approaches infinity and the numerator approaches zero, the limit of the function is zero as x approaches infinity.  

Normally, you would factor out x-9 from the numerator and denominator, but it isn't necessary for this problem.  Since the problem asks for the limit to be evaluated at x=5 where there is no discontinuity, the limit will be f(5).

For evaluating limits, we need to consider what happens to the function when we input larger and larger values.  In this case, the numerator's leading exponent is higher than the leading exponent than the denominator.  Therefore, the numerator will be growing at a faster rate than the numerator and always be a larger value.  Therefore:

Evaluating infinity limits is done by simply considering the leading term in both the numerator and denominator (all of the other parts of the equation will become insignificant when we plug in more and more negative values).  In this case, the leading terms have the same exponents.  Therefore, the limit is just the ratio of the leading coefficients:

For infinity limits, we only need to consider the leading terms in both the numerator and the denominator.  Everything else will become insignificant at the end behavior. In this case, the denominator has a larger leading exponent than the denominator.  Therefore, the denominator is getting larger than the numerator.  This type of function always tends to zero:

To solve this problem one must first note that upon direct substitution of , the function is undefined and therefore, the limit cannot be evaluated through direct substitution. However, use of L'Hospital's Rule significantly helps, to do this one evaluates the limit at the same point but first must take the derivative of both the numerator and denominator. 

Our problem now becomes:
 

So just because the left hand limit agrees with the right hand limit does not necessarily mean that the function is continuous at that point. There can be a point that is called a removable discontinuity, is not continuous at that point, even though that limit of that point does exist. In addition, that discontinuous point would not be in the domain of the function. 

Think of the limit as tracing pencils along the function towards the point the limit is being evaluated at. With the limit, you will never touch the point you are going towards, you are just looking for the value that you are tending towards as you reach some point x=a. When you plug x=a into the function though, you are finding the value at that exact point. So you can have a limit exist with no value at the exact point. 

This question is asking to evaluate a one-sided equation of a function. Specifically, the limit of the function to the left of . When  is substituted into the function the result is indeterminate. This means it is in the form zero over zero. 

Since the question is looking for a one-sided limit, let us substitute in a value that is slightly less than .