For the function A whose graph is shown, state the following. (If the limit is infinite, enter '∞' or '-∞', as appropriate. If the limit does not otherwise exist, enter DNE.)The x y-coordinate plane is given. The function enters the window in the second quadrant, goes up and right becoming more steep, exits just to the left of x = −3 in the second quadrant nearly vertical, reenters just to the right of x = −3 in the second quadrant nearly vertical, goes down and right becoming less steep, crosses the x-axis at x = −2, goes down and right becoming more steep, exits the window just to the left of x = −1 in the third quadrant nearly vertical, reenters just to the right of x = −1 in the third quadrant nearly vertical, goes up and right becoming less steep, crosses the y-axis at approximately y = −0.6, changes direction at the approximate point (0.5, −0.5) goes down and right becoming more steep, exits the window just to the left of x = 2 in the fourth quadrant nearly vertical, reenters just to the right of x = 2 in the first quadrant nearly vertical, goes down and right becoming less steep, crosses the x-axis at x = 3, changes direction at the approximate point (4.5, −1.5), goes up and right becoming more steep, crosses the x-axis at approximately x = 6.5, and exits the window in the first quadrant.(a)lim x → −3 A(x) (b)lim x → 2− A(x) (c)lim x → 2+ A(x) (d)lim x → −1 A(x) (e)The equations of the vertical asymptotes. (Enter your answers as a comma-separated list.)
Question
For the function A whose graph is shown, state the following. (If the limit is infinite, enter '∞' or '-∞', as appropriate. If the limit does not otherwise exist, enter DNE.)The x y-coordinate plane is given. The function enters the window in the second quadrant, goes up and right becoming more steep, exits just to the left of x = −3 in the second quadrant nearly vertical, reenters just to the right of x = −3 in the second quadrant nearly vertical, goes down and right becoming less steep, crosses the x-axis at x = −2, goes down and right becoming more steep, exits the window just to the left of x = −1 in the third quadrant nearly vertical, reenters just to the right of x = −1 in the third quadrant nearly vertical, goes up and right becoming less steep, crosses the y-axis at approximately y = −0.6, changes direction at the approximate point (0.5, −0.5) goes down and right becoming more steep, exits the window just to the left of x = 2 in the fourth quadrant nearly vertical, reenters just to the right of x = 2 in the first quadrant nearly vertical, goes down and right becoming less steep, crosses the x-axis at x = 3, changes direction at the approximate point (4.5, −1.5), goes up and right becoming more steep, crosses the x-axis at approximately x = 6.5, and exits the window in the first quadrant.(a)lim x → −3 A(x) (b)lim x → 2− A(x) (c)lim x → 2+ A(x) (d)lim x → −1 A(x) (e)The equations of the vertical asymptotes. (Enter your answers as a comma-separated list.)
Solution
(a) Para encontrar el límite cuando se aproxima a de la función , observamos el comportamiento de la gráfica cerca de . La función se vuelve casi vertical a la izquierda y a la derecha de , lo que sugiere la presencia de una asíntota vertical. Por lo tanto, el límite no existe (DNE).
(b) Para encontrar el límite cuando se aproxima a por la izquierda (), observamos que la función se vuelve casi vertical justo a la izquierda de en el cuarto cuadrante. Esto sugiere que la función tiende a . Por lo tanto, .
(c) Para encontrar el límite cuando se aproxima a por la derecha (), observamos que la función se vuelve casi vertical justo a la derecha de en el primer cuadrante. Esto sugiere que la función tiende a . Por lo tanto, .
(d) Para encontrar el límite cuando se aproxima a de la función , observamos el comportamiento de la gráfica cerca de . La función se vuelve casi vertical a la izquierda y a la derecha de , lo que sugiere la presencia de una asíntota vertical. Por lo tanto, el límite no existe (DNE).
(e) Las ecuaciones de las asíntotas verticales se encuentran en los valores de donde la función se vuelve casi vertical y el límite no existe. De la descripción, estos valores son y . Por lo tanto, las ecuaciones de las asíntotas verticales son .
Similar Questions
For the function f whose graph is shown, state the following. (If the limit is infinite, enter '∞' or '-∞', as appropriate. If the limit does not otherwise exist, enter DNE.)FigureThe x y coordinate plane is given. Refer to the adjacent description for more details.Description(a)lim x → −7 f(x)(0,2) (b)lim x → −3 f(x)DNE (c)lim x → 0 f(x)2 (d)lim x → 6− f(x)∞ (e)lim x → 6+ f(x)−∞ (f)the equations of the vertical asymptotes (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)x = (−3,6)
The graphs of f and g are given. Use them to evaluate each limit, if it exists. (If an answer does not exist, enter DNE.)The x y-coordinate plane is given. A function with four parts, labeled y = f(x), is shown.The first part is a curve, enters the window in the second quadrant, goes down and right becoming less steep, crosses the x-axis at x = −2, and ends at the point (−1, −1).The second part is linear, starts at the point (−1, −1), goes up and right, passes through the x-axis at x = 0, and ends at the open point (2, 2).The third part is a closed point located at (2, 1).The last part is linear, starts at the open point (2, 2), goes down and right, and exits the window in the first quadrant. The x y-coordinate plane is given. A function with two parts, labeled y = g(x), is shown.The first part is a curve, enters the window in the third quadrant, goes up and right becoming less steep, crosses the x-axis at x = −1, crosses the y-axis at approximately 1.3, and ends at the open point (1, 2).The second part is a curve, begins at the closed point (1, 1), goes down and right becoming more steep, crosses the x-axis at x = 2, and exits the window in the fourth quadrant.(a)lim x→2 [f(x) + g(x)] (b)lim x→1 [f(x) + g(x)] (c)lim x→0 [f(x)g(x)] (d)lim x→−1 f(x)g(x) (e)lim x→2 [x3f(x)] (f)lim x→1 3 + f(x)
The graph of a function f is given in the figure.A curve is shown on the x y coordinate plane. It begins at the point (−2, −1), goes up and to the right, passes through the approximate point (−1, −0.2), and passes through the negative x-axis at the approximate point (−0.8, 0). It then continues up and right, passes through the positive y-axis at the point (0, 1), and reaches a high point at (1, 3). It then goes down and right, passes through the points (2, 2) and (3, 1), and ends at the approximate point (4, 0.5).(a)Find the value of f(1).(b)Estimate the value of f(−1).(c)For what values of x is f(x) = 1? (Enter your answers as a comma-separated list.) (d)Estimate the value of x such that f(x) = 0.x = (e)State the domain and range of f. (Enter your answers in interval notation.)domain range (f)On what interval is f increasing? (Enter your answer using interval notation.)
The graph of a function g is shown.The x y-coordinate plane is given. A function labeled y = g(x) with 4 parts is graphed.The first part is a curve, enters the window in the second quadrant, goes up and right becoming less steep, crosses the y-axis at approximately y = 2.5, and ends at the open point (2, 3).The second part is a curve begins again at the open point (2, 1), goes up and right becoming less steep, and ends at the open point (5, 2).The third part is the closed approximate point (5, 1.2).The fourth part is a curve, begins at the open point (5, 2) goes down and right becoming more steep, and exits the window in the first quadrant.Use it to state the values (if they exist) of the following:(a) lim x → 2− g(x)(b) lim x → 2+ g(x)(c) lim x → 2 g(x)(d) lim x → 5− g(x)(e) lim x → 5+ g(x)(f) lim x → 5 g(x)SolutionLooking at the graph we see that the values of g(x) approach as x approaches 2 from the left, but they approach as x approaches 2 from the right.Therefore (a) lim x → 2− g(x) = and (b) lim x → 2+ g(x) = .Since the left and right limits are different, we conclude that (c) the limit as x approaches 2 of g(x) does not exist.The graph also shows that (d) lim x → 5− g(x) = and (e) lim x → 5+ g(x) = .This time, the left and right limits are the same and so, by this theorem, we have (f) lim x → 5 g(x) = Despite this fact, notice that g(5) ≠ 2.
Describe the behaviour of the graph of the function as x → – ∞ and as x → ∞.
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