Evaluate the limit: lim(x→3) (2x² − 5x + 1).
- 4 Correct answer.
- 10
- 22
- 16
Explanation
A polynomial is continuous everywhere, so the limit equals the function value at x = 3. Substituting gives 2(9) − 15 + 1 = 18 − 15 + 1 = 4.
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A polynomial is continuous everywhere, so the limit equals the function value at x = 3. Substituting gives 2(9) − 15 + 1 = 18 − 15 + 1 = 4.
The expression is 0/0 at x = 2, so factor the numerator as (x − 2)(x + 2). Canceling the common factor leaves x + 2, whose limit as x→2 is 4.
When numerator and denominator have the same degree, the limit at infinity is the ratio of leading coefficients. Here that ratio is 3/1 = 3.
Using the standard limit lim(u→0) sin(u)/u = 1 with u = 5x, rewrite the expression as 5 · (sin 5x)/(5x). The limit is therefore 5 · 1 = 5.
The limit describes the behavior of f near x = 4, not at x = 4 itself. A limit of 7 means the function values approach 7 as x approaches 4, regardless of the function's value there.
As x approaches −2 from the right, x + 2 is small and positive, so the reciprocal grows without bound in the positive direction. Hence the one-sided limit is +∞.
Multiply numerator and denominator by the conjugate √(x + 9) + 3 to get x/(x(√(x + 9) + 3)). Canceling x leaves 1/(√(x + 9) + 3), which tends to 1/(3 + 3) = 1/6.
The numerator and denominator both have degree 3, so the limit at infinity equals the ratio of leading coefficients, 4/2 = 2. Lower-degree terms become negligible.
The limit as x→1 of (x² − 1)/(x − 1) equals 2, but f(1) = 3, so the limit exists yet differs from the function value. This is a removable discontinuity because redefining f(1) as 2 would make it continuous.
Multiply numerator and denominator by 1 + cos x to obtain (sin²x)/(x²(1 + cos x)). Since (sin x)/x → 1 and 1 + cos x → 2, the limit is 1/2.
The limit of a product is the product of limits, giving 5 · (−2) = −10, and the limit of the sum adds lim f = 5. Thus −10 + 5 = −5. Wait-recompute: the expression is f·g + f = −10 + 5 = −5.
Both numerator and denominator vanish at x = 2, so factor: numerator (x − 2)(x + 2), denominator (x − 2)(x − 1). Canceling x − 2 leaves (x + 2)/(x − 1), which tends to 4/1 = 4.
Continuity at x = 1 requires f(1) to equal the limit as x→1. Since 2 ≠ 4, the function fails the continuity condition and is discontinuous there.
Multiply by the conjugate over itself: (x² + 3x − x²)/(√(x² + 3x) + x) = 3x/(√(x² + 3x) + x). Dividing by x gives 3/(√(1 + 3/x) + 1), which tends to 3/2.
Both bounding functions tend to 2 as x→0, since x² → 0. Because g is trapped between two functions with the same limit, the Squeeze Theorem gives lim g(x) = 2.
As x approaches 0 from the left, 1/x → −∞, and e raised to a quantity tending to −∞ approaches 0. Hence the left-hand limit is 0.
Continuity at x = 2 requires the left-hand limit 2a + 1 to equal the right-hand value 2² − 3 = 1. Solving 2a + 1 = 1 gives a = 0. Wait-this yields a = 0. Rechecking: 2a + 1 = 1 gives a = 0.
Write tan x = (sin x)/(cos x), so (tan x)/x = [(sin x)/x] · [1/(cos x)]. As x→0, the first factor tends to 1 and the second tends to 1, giving a limit of 1.
Factor both: x³ − 1 = (x − 1)(x² + x + 1) and x² − 1 = (x − 1)(x + 1). Canceling x − 1 leaves (x² + x + 1)/(x + 1), which at x = 1 equals 3/2.
Because |g(x)| ≤ 5, we have |f(x)g(x)| ≤ 5|f(x)|, and 5|f(x)| → 0. By the Squeeze Theorem, the product tends to 0 regardless of the behavior of g beyond being bounded.
Factor the numerator as (3x - 2)(x - 1), cancel the common factor (x - 1), leaving 3x - 2. Substituting x = 1 gives 3(1) - 2 = 1. The limit exists even though the expression is undefined at x = 1.
The power rule states that the derivative of x^n is n·x^(n-1). Applying it with n = 3 gives 3x^2.
Continuity at a point requires three things: f(a) is defined, the limit exists, and the limit equals f(a). Only the third condition is listed among the choices.
The standard derivative formula for the sine function is d/dx[sin x] = cos x. This follows from the limit definition using the special limits for sine and cosine.
Rewrite the expression as 4·(sin(4x))/(4x). Since (sin u)/u approaches 1 as u approaches 0, the limit is 4·1 = 4.
By the chain rule, d/dx[e^u] = e^u·u'. With u = 2x, u' = 2, so the derivative is 2e^(2x).
A horizontal tangent means f'(c) = 0, a critical point. If the second derivative is positive there, the graph is concave up, so the critical point is a local minimum by the second derivative test.
The derivative at a point is defined as the limit of the difference quotient (f(a+h) - f(a))/h as h approaches 0. This limit, when it exists, is the tangent slope.
The product rule gives f'(x) = (x^2)'·cos x + x^2·(cos x)' = 2x·cos x + x^2·(-sin x). Simplifying yields 2x·cos x - x^2·sin x.
By the quotient rule, the derivative is [(3)(x - 2) - (3x + 1)(1)]/(x - 2)^2. The numerator simplifies to 3x - 6 - 3x - 1 = -7, giving -7/(x - 2)^2.
The chain rule gives dy/dx = 4(2x + 5)^3·(2), since the inner derivative of 2x + 5 is 2. Multiplying the constants yields 8(2x + 5)^3.
Velocity is s'(t) = 3t^2 - 12t + 9 = 3(t - 1)(t - 3). The particle is at rest when velocity is zero, which occurs at t = 1 and t = 3.
Using the chain rule, d/dx[ln(5x)] = (1/(5x))·5 = 1/x. This matches the fact that ln(5x) = ln 5 + ln x, whose derivative is 1/x.
Differentiating both sides with respect to x gives 2x + 2y·(dy/dx) = 0. Solving for dy/dx yields dy/dx = -2x/(2y) = -x/y.
Compute f''(x) = 6x. The graph is concave down where f''(x) < 0, which means 6x < 0, or x < 0. Thus the interval is (-infinity, 0).
Factor both parts: (x - 3)(x + 3) over x(x - 3). Cancel (x - 3) to get (x + 3)/x, then substitute x = 3 to obtain 6/3 = 2.
The derivative y' = 2x equals 4 at x = 2, so the tangent slope is 4. Using point-slope form, y - 4 = 4(x - 2), which simplifies to y = 4x - 4.
Apply the product rule: f'(x) = (1)·e^x + x·e^x = e^x(1 + x). This factors as (x + 1)e^x.
The Mean Value Theorem guarantees some c in (1, 4) with f'(c) equal to the average rate of change (f(4) - f(1))/(4 - 1) = (9 - 3)/3 = 2.
The standard derivative of the inverse tangent is 1/(1 + x^2). It comes from implicit differentiation of tan y = x, giving sec^2(y)·y' = 1 and y' = 1/(1 + x^2).
Critical points satisfy f'(x) = 3x^2 - 6x = 3x(x - 2) = 0, giving x = 0 and x = 2. Evaluating f at critical points and endpoints: f(-1) = 0, f(0) = 4, f(2) = 0, f(4) = 20. The largest is 20.
The difference quotient is (sqrt(x + h) - sqrt(x))/h. Multiplying numerator and denominator by the conjugate gives 1/(sqrt(x + h) + sqrt(x)), whose limit as h approaches 0 is 1/(2·sqrt(x)).
Differentiate implicitly: 2x + y + x·y' + 2y·y' = 0. Solving gives y' = -(2x + y)/(x + 2y). At (1, 2) this is -(2 + 2)/(1 + 4) = -4/5.
Let x be the base distance and y the height, so x^2 + y^2 = 100. Differentiating gives 2x·x' + 2y·y' = 0. When x = 6, y = 8, so y' = -(6·2)/8 = -3/2, meaning the top descends at 3/2 ft/s.
A sign change in f'' at a point where the function is defined indicates a change in concavity, which defines an inflection point. The second derivative test is inconclusive when f''(3) = 0.
The derivative of any constant function is zero because the graph is a horizontal line with slope 0 everywhere.
The derivative of sine is cosine, a standard derivative rule: d/dx[sin x] = cos x.
The power rule states that d/dx[x^n] = n x^(n-1). Applying it with n = 5 gives 5x^4.
Differentiate term by term using the power rule: d/dx[3x^2] = 6x, d/dx[2x] = 2, and d/dx[-1] = 0, giving 6x + 2.
The derivative of the natural logarithm is the reciprocal function: d/dx[ln x] = 1/x for x > 0.