Відео

If you look at the graph of cot(x) ⁡ , you will see it is a periodic function, going from positive infinity to negative infinity indefinitely. No matter how large the value of x gets, it will keep repeating this way and never tends to any one value of y . Hence, there is no limit.

@@hutchmath thank you

trust me even as a math major theres some things u just wont get for too long😂

The algebraic processes will be the same. The only time there is a difference is if you have division by zero and there is nothing you can do algebraically to prevent division by zero. In this case the limit will go to either infinity or negative infinity.

@@hutchmath Thanks for writing back. Even in that case, the algebraic calculations will be the same in the case of of one-sided vs. normal (from both sides) limits. It is simply that in order to see if a normal limit is approaching infinity, we would calculate both + and - and see if they both match. Please correct me if I am wrong, thanks again!

The only time a one-sided limit will not exist is if you have division by zero or if you are trying to take the even root of a negative number, such as a square root. For example, in the problem I did in this video with sqrt(16-x^2), the limit as we approach 4 from the positive side will not exist. This is because any number larger than 4 will cause a negative number under the square root. The domain for this problem is not defined for numbers larger than 4, which is why the righthand limit does not exist. For our second example, we were able to use algebraic methods to avoid division by zero, so this limit exists. However, if we are trying to find the one-sided limit at -2, this would not exist since we would still have division by zero even after factoring and cancelling. Therefore, if you do not have division by zero or the even root of a negative number, then the one-sided limit will exist. To see if a regular limit exist (not one-sided), the left and right hand limits must both exist at that number.

@@hutchmath Thanks a lot sir! I am learning calculus by my own!! Was having problem with this thing! Thanks a lot for such a clear explanation! Pray for me so that I can complete calculus to start with calculus based physics

So yes, the area is used as a constraint here, which is why we used the starting equation xy=50. However, if we want to minimize the amount of paper used, this means we need to minimize the area of the entire poster including the margins. We would not use perimeter here as the objective equation.

With our table of values and our graph, we can determine the vertex, which would be at (0,4). We can use the vertex form of a quadratic, which is y = a(x-h)^2+k. Since our vertex is at (0,4), this means that h=0 and k=4. We can put this into the equation to get y = a(x-0)^2+4. Now we need to find the "a" value. To do this, pick any other point on our table and put this into our equation for x and y. Let's use the point (1, 0) from our table. Then x = 1 and y = 0. Plug this into our equation to get: 0 = a(1-0)^2+4. This simplifies to 0 = a+4. Subtract the 4 to get a = -4. Therefore, our equation is y = -4(x-0)^2+4 which simplifies to y = -4x^2 + 4.