First, try evaluating the limit at the target value.

This gives us an indeterminate form, so we have to keep trying. Let's factor the polynomials:

We can cancel an , so let's do that.

Now evaluate at the target value.

The limit evaluates to .

The mid-point Reimann sum is given by this formula:

, where , and  is the number of 

intervals.  Thus, if the region from  to  is divided into twenty intervals,  and .  For ten intervals,  and .  For five internals,  and .  The higher the number of intervals, the more precise the estimation.  Thus, when  (and hence ), the estimation is the most accurate.  

For this value, the Limit Laws can be applied: 

Removable discontinuities don't affect the limiting process. The limit process is essentially saying "As you get arbitrarily close to , the function is getting arbitrarily close to ." As you can see, whether or not the function is defined at  is irrelevant, because we're want to look at values close to it, but never exactly at .

Graphically, we can see that the left and right limits are both three. If you trace the graph from either the left or right of , you will end up at .

By definition, if both the right and left limit evaluate to the same thing, the actual limit must agree. It's not possible for both of them to be three, but for the limit to disagree with them (at least in a two dimensional graph!)

Thus the correct answer is that the limit exists, and is three.

A right limit is found graphically by starting at a point to the right of the specified x value and tracing along the graph to the left until you hit the specified x. Regardless of whether or not the function is defined, or perhaps if its defined, but not where your finger ends up, your finger should be at the limit. This is because limits don't concern themselves with what happens at the specified x value. For right limits, we're essentially asking "If I get arbitrarily close to  from the right, what is  getting close to?"

The same holds true for left limits, but you approach them from the other side.

In this problem,  is a right limit. Following the above method, we find that as we approach  from the right, we end up at . Similarly,  is a left limit. Approaching  from the left, we see that we arrive at .

Lastly, a regular limit is defined when both the right and left limits are defined and equal. In this case, the right limit is 5 and the left limit is 3, so the regular limit does not exist 

The derivative of the function y = sec(x) is sec(x)tan(x). First take the derivative of the outside of the function: y = sec(4x³) : y' = sec(5x³)tan(5x³). Then take the derivative of the inside of the function: 5x³ becomes 15x². So your final answer is: y' = ec(5x³)tan(5x³)15x²

First find the derivative of the function.

f(x) = –x² + 5√(x)

f'(x) = –2x + 5(1/2)x^–1/2

Simplify the problem

f'(x) = –2x + (5/2x^1/2)

Plug in 9. 

f'(3) = –2(9) + (5/2(9)^1/2)

= –18 + 5/(6)

Rewrite problem. 

(x + 1)/(x – 1)

Use quotient rule to solve this derivative. 

((x – 1)(1) – (x + 1)(1))/(x – 1)²

(x – 1) – x – 1)/(x – 1)²

–2/(x – 1)²

Use the chain rule and the formula

The answer is . It is easy to solve if we multiply everything together first before taking the derivative.