Hi, I need help with my calculus homework, thank you.
A student writes the following: 422+13 Problem: Evaluate the limit Iim $_2 23}2 Student answer: Line 1: Direct Substitution is not possible because the denominatOI is 0 when a: = 2 . Line 2: Instead let's multiply by the conjugate (to rationalize). . . 43+13 . ( 4m+13)( 4$+1+3) L 3: = : Line 4' 2 lm l4$+ll_g ' ac>2 \"72 Line 5:: lm 4l$_2l $>2 332 Line 6: : lm 4 : 4 $>2 Is the student's work correct? (Write "yes" or "no".) If incorrect, in what line is the first error? (Write 1, 2, 3, 4, 5, or 6. If it is correct, write "none". If there is more than one error, write the line of the FIRST error.) The correct final answer to the problem isIs it true that the following is another way of writing the definition? A function f (a) is continuous at a point a if and only if all three of f (a) lim f (a) lim f (x) x -at exist and are equal (that means f (a) = lim f (x) = lim f (x)) x - A function is discontinuous at a point a if it fails to be continuous at a . Is it true that the above is another (equivalent) way of writing the definition? O True O FalseConsider the graph of f (a): 0 2 6 At x = -1 the function is/has a [Select ]At a: = 1 the function isfhas a ' [59'9\"] V '. At a: : 0 the function is/has 3' [59'9\"] V'. At a: = 1 the function is/has a' [59'9\"] V'. At a: : 2 the function is/has 3 [59'9\"] V . At a: = 3 the function is/has a [59'9\"] V . At a: : 4 the function is/has 3 [59'9\"] V . At a: = 5 the function is/has a [59'9\"] V . At a: : 6 the function is/has 3 [59'9\"] V . 3:2 + 1 if :1: 2 We aim to nd out if f (:3) has a discontinuity at a: = 2, and if so, of what type. In order to do that, rst nd the following information: f(2) = [Select] v H111 fer) = [Select] v a:>2_ ljjll f0\") = [Select] v at>2+ Is f (3:) continuous or discontinuous at a: = 2 ? [ Select] v If f (3:) is discontinuous at a: : 2 , what type of discontinuity is it? [ Select] v Suppose we know the following information about the function f (x) : f (-1) = -4, f(2.5) =3, f() = 2.4 and f (1) does not exist limx -1- f (2) = -4 lim -1+ f (2) = -4 limx +2.5+ f (2) = -00 lima-T f (x) = 0 limx +8- f (x) = 3 limx 8+ f (x) = 3.01 What does this information tell us about the continuity of f (a) ? At a = -1, f (a) is/has a [ Select ] At a = 1, f (a) is/has a [ Select ] At a = 2.5, f (a) is/has a [ Select ] At a = 1, f (a) is/has a [ Select ]At x - 8, f (x) is/has a [ Select ]Which of the following statements are TRUE? (Select all that apply.) O All polynomials are continuous at every point in their domain O All polynomials are continuous at every point in R. O The function f (a) = 5x+4 12 -3 is a rational function, and it is continuous at every point in R. All trigonometric functions are continuous at every point in their domain All piecewise defined functions are continuous at every point in their domain O All rational functions are continuous at every point in their domainWhich of the following statements are TRUE? a} If f (m) is continuous on the interval [0, 5] and f(0) = 1 and f (5) = 2, then the Intermediate Value Theorem says that there is a numberc in [0, 5] such that f(c) : [Select] b) If f (3:) is continuous on the interval [2, 4] and f (2) 0 , then the Intermediate Value Theorem says that f (3) = 0. [ Select] v c) If f (a?) is any function and f (a) = 4 and f(b) = 6, then the Intermediate Value Theorem says that there is a number 0 in [0,, b] satisfying f(c) : 5. [Select] v
