Limits and Continuity: Calculus 1 Homework Problems Part 5

In the live stream we will go over Homework Problems on limits and continuity. We will solve more than 30 problems. Please do not forget to like the live stream and subscribe to our channel. Thanks! Let us solve the following limit problem. We are asked to compute the limit as x goes to 2 of the fraction, x minus, 2 over the quantity square root of x square , + 60 minus 8. We will show you how to solve this problem using limits laws, and basic algebraic techniques such as the factorization of polynomial, and the rationalization of irrational expression using their conjugates. We first define the function f of x as the fraction, x minus, 2 over square root of x square , + 60 minus 8 We first try to compute f of 2. When we compute f of 2, we easily see that f of 2 is 0 over 0. This is an indeterminate form. In order to address this issue we have to simplify the whole expression using basic algebraic techniques. This implies that, we have to multiply both the numerator and denominator with the conjugate of square root of x square , + 60 minus 8 which is square root of x square , + 60 + 8. So, we have to expand the denominator using properties of the square root. The new expression is x minus, 2 times the square root of x square , + 60 + 8 x minus, 16 over, square root of x square , + 60 minus 8 times, square root of x square , + 60 + 8 After simplification, the new expression is, x minus, 2, times square root of x square , + 60 + 8, over x square minus, 4.We obviously see that we can factorize the numerator and the denominator. After factorization, we see that the common factor which appears on both the numerator and denominator is x minus, 2 We simplify the common factor, and obtain; the limit as x goes to 2 of 8 + square root of x square , + 60 over, x , + 2. Finally, when we plug 2 in the new expression we find out that the limit is 16 over 4 ##NewProb## In problem 2, we will study the continuity of the piecewise function g of x, defined as follows; g of x is 3 x cube , + 11 x square, + 9 x, + 9, over negative x minus 3 for x not equal to negative 3, and g of x is 7 x + 8 for x equal to negative 3. Our goal is to determine if g of x is continuous at the point x = negative 3. If it is not continuous, we will determine the type of discontinuity it has. Recall that the function g of x is continuous at x = negative 3 if and only if the limit of g of x as x approaches negative 3 is equal to g of negative 3. We should start by evaluating the limit of the function as x approaches negative 3. Let us compute the limit as x goes to negative 3 of the fraction 3 x cube , + 11 x square, + 9 x, + 9, over the polynomial negative x minus 3. It is better to start by finding the value of the fraction at the point x = negative 3. After evaluating the fraction at the point negative 3, we have minus 81, plus 99 minus 27, plus 9, over 3 minus 3, which equals 0 over 0. We obtain an indeterminate form of the type 0 over 0. Therefore, we have to factorize the whole fraction in order to simplify the common factor. This sometimes happens when the quantity x plus 3 is a common factor of both the numerator and the denominator The polynomial 3 x cube , + 11 x square, + 9 x, + 9 is divisible by x, + 3, since they have a common root, -3. When we divide 3 x cube , + 11 x square, + 9 x, + 9 by x, + 3, we get 3 x square , + 2 x, + 3. Therefore, 3 x cube , + 11 x square, + 9 x, + 9 can be written as the product of x, + 3 and 3 x square , + 2 x, + 3. We notice that the numerator can be factorized as the polynomial x, + 3) times the polynomial 3x square, + 2x, + 3). and the denominator can be factorized as the polynomial x, + 3) times the polynomial minus 1). The common factor in both expressions that should be simplified is x, + 3. by replacing negative 3 in this new function, we find that the limit is, minus 24 . The limit of the function g of x as x approaches -3 is, minus 24 . The value of the function g at x = -3 is -13. The function is discontinuous at x = -3 and has a removable discontinuity. This is because the limit does not equal the value of the function at x = -3. ##NewProb## In Problem 3, we will study the continuity of the piecewise function g of x, defined as follows; g of x is negative 2 x cube , + 11 x square, + 25 x minus 28, over x square minus 14 x, + 49 for x less than 7, and g of x is, minus 8 x for x greater or equal to 7. We will try to find if g of x is continuous at the point x = 7. If it is not continuous, we will determine the type of discontinuity it has. The function g of x is continuous at x = 7 if and only if the limit of g of x as x approaches 7 is equal to g of 7. We will begin by evaluating the limit of the function as x approaches 7. The limit of the function exists if and only if the left-hand limit and the right-hand limit exist and are equal. We will now evaluate the left-han