# Calculus 2 : Limits

## Example Questions

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### Example Question #1 : Calculus Ii

Evaluate:

None of the other answers is correct.

The limit does not exist.

None of the other answers is correct.

Explanation:

Evaluated at , the numerator and the denominator are both equal to 0, as shown below:

So a straightforward substitution will not work. L'Hospital's rule will work here, but an easier way is to note that

and .

So the expression can be rewritten - and solved - as follows:

### Example Question #1 : Limit Concepts

Evaluate the limit:

Does Not Exist

Explanation:

Directly evaluating the limit will produce an indeterminant answer of .

Rewriting the limit in terms of sine and cosine, , we can try to manipulate the function in order to utilize the property .

Multiplying the function by the arguments of the sine functions, , we can see that the limit will be .

### Example Question #1 : Limits

Evaluate .

The limit does not exist.

Explanation:

and

,

so we cannot solve this by substituting.

However, we can rewrite the expression:

### Example Question #4 : Calculus Ii

Find the limit of  as  approaches infinity.

Inconclusive

Explanation:

The expression  can be rewritten as .

Recall the Squeeze theorem can be used to solve for the limit.  The sine function has a range from , which means that the range must be inside this boundary.

Multiply the  term through.

Take the limit as  approaches infinity for all terms.

Since the left and right ends of this interval are zero, it can be concluded that  must also approach to zero.

### Example Question #1 : Limit Concepts

Determine the limit.

Explanation:

To determine, , graph the function  and notice the direction from the left and right of the curve as it approaches .

Both the left and right direction goes to negative infinity.

### Example Question #6 : Calculus Ii

Which of the following is true?

If neither  nor  exist, then  also doesn't exist.

If  exists, then  and  both exist.

and  exist if and only if  exists.

If  and , then  exists.

If  and , then  exists.

Explanation:

If  and , then  exists.

This can be proven rigorously using the  definition of a limit, but it is most likely beyond the scope of your class.

### Example Question #7 : Calculus Ii

Determine the limit:

Explanation:

Isolate the constant in the limit.

The limit property .

Therefore:

### Example Question #8 : Calculus Ii

Evaluate the limit, if possible:

Explanation:

To evaluate , notice that the inside term  will approach infinity after substitution.  The inverse tangent of a very large number approaches to .

### Example Question #9 : Calculus Ii

Evaluate the following limit:

Explanation:

The first step is to factor out the highest degree term from the polynomial on top and bottom (essentially pulling out 1):

which becomes

Evaluating the limit, we approach .

### Example Question #10 : Calculus Ii

Evaluate the following limit:

Explanation:

To evaluate the limit, first pull out the largest power term from top and bottom (so we are removing 1, in essence):

which becomes

Plugging in infinity, we find that the numerator approaches zero, which makes the entire limit approach 0.

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