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Recognizing the difference quotient for the function f(t) = e^(2t) at t = 0, the limit is f'(0). Since f'(t) = 2 e^(2t), we find f'(0) = 2 * e^0 = 2.
Applying the product rule (uv)' = u'v + uv' with u(x) = 3x + 1 and v(x) = e^(-x), we obtain u'(x) = 3 and v'(x) = -e^(-x). Thus f'(x) = 3 e^(-x) - (3x + 1) e^(-x) = (3 - 3x - 1) e^(-x) = (2 - 3x) e^(-x).
The homogeneous linear differential equation y' + ay = 0 has general solution y(x) = C e^(-ax). Here a = 2, so y(x) = C e^(-2x). Applying y(0) = 5 yields C = 5, giving y(x) = 5 e^(-2x).
An antiderivative of 2x + 3 is F(x) = x^2 + 3x. Evaluating between 0 and 1 gives F(1) - F(0) = (1^2 + 3 * 1) - 0 = 4.
The first derivative is f'(x) = 3x^2 - 12x + 9, so the second derivative is f''(x) = 6x - 12. Setting f''(x) = 0 yields x = 2. Since f''(x) changes sign at x = 2, an inflection point occurs at x = 2. Evaluating f(2) = 2^3 - 6(2^2) + 9(2) + 2 = 8 - 24 + 18 + 2 = 4, so the point is (2, 4).
The equation of the tangent line at x = a is y = f'(a)(x - a) + f(a). For f(x) = ln(x) at a = 1, f(1) = 0 and f'(x) = 1/x so f'(1) = 1. Thus y = 1*(x - 1) + 0, which simplifies to y = x - 1.
f(x) matches the pattern u'(x)/u(x) where u(x) = x^2 + 1 > 0. The general antiderivative is F(x) = ln(x^2 + 1) + C. Imposing F(0) = ln(1) + C = C = 3 yields F(x) = ln(x^2 + 1) + 3.
For a composite function ln(u(x)), the chain rule gives [ln(u(x))]' = u'(x) / u(x). Here u(x) = x^2 + 4, so u'(x) = 2x. Thus g'(x) = 2x / (x^2 + 4).
A constant particular solution is y_p = -6 / (-3) = -2 (or testing y' = 0 gives -3y = 6 => y = -2). The general solution to the homogeneous equation y' - 3y = 0 is y_h(x) = C e^(3x). Thus y(x) = C e^(3x) - 2. Applying y(0) = 1 gives C - 2 = 1 => C = 3, so y(x) = 3 e^(3x) - 2.
Notice that x e^(x^2) = (1/2) * (2x e^(x^2)) = (1/2) * [e^(x^2)]'. Thus an antiderivative is F(x) = (1/2) e^(x^2). Evaluating from 0 to 2 yields F(2) - F(0) = (1/2)(e^(2^2) - e^0) = (e^4 - 1) / 2.
First derivative: h'(x) = 1*e^(2x) + x*(2e^(2x)) = (2x + 1)e^(2x). Second derivative: h''(x) = 2*e^(2x) + (2x + 1)*(2e^(2x)) = (4x + 4)e^(2x) = 4(x + 1)e^(2x). Since e^(2x) > 0 always, h''(x) >= 0 if and only if x + 1 >= 0, i.e., x >= -1. Hence h is convex on [-1, +inf[.
Multiplying the numerator and denominator by the conjugate (sqrt(x + 4) + 2) gives [(x + 4) - 4] / [x (sqrt(x + 4) + 2)] = x / [x (sqrt(x + 4) + 2)] = 1 / (sqrt(x + 4) + 2). As x -> 0, this evaluates to 1 / (sqrt(4) + 2) = 1 / (2 + 2) = 1/4.
Notice the structure u'(x) u(x) where u(x) = ln(x) and u'(x) = 1/x. An antiderivative is (1/2) (ln x)^2. Evaluating between 1 and e: (1/2) [ (ln e)^2 - (ln 1)^2 ] = (1/2) [ 1^2 - 0 ] = 1/2.
Using the Taylor expansion of e^x around 0: e^x = 1 + x + x^2 / 2! + o(x^2). Substituting gives e^x - x - 1 = x^2 / 2 + o(x^2). Dividing by x^2 yields 1/2 + o(1), which approaches 1/2 as x -> 0.
Rewrite f(x) as e^(x ln x). Differentiating using the chain rule: f'(x) = [x ln x]' e^(x ln x). By the product rule, [x ln x]' = 1*ln(x) + x*(1/x) = ln(x) + 1. Therefore, f'(x) = (ln(x) + 1) x^x.
An antiderivative of 1 / (1 + x^2) is arctan(x). Evaluating between 0 and 1 gives arctan(1) - arctan(0) = pi / 4 - 0 = pi / 4.
Notice that sin(x) cos(x) = (1/2) sin(2x), or use the substitution u = sin(x) where du = cos(x) dx. The integral becomes int (0 to 1) u du = [u^2 / 2]_0^1 = 1/2.
The characteristic equation is r^2 + 4 = 0, giving roots r = +- 2i. The general real solution is y(x) = A cos(2x) + B sin(2x). The initial condition y(0) = 0 gives A = 0. The derivative is y'(x) = 2B cos(2x). Imposing y'(0) = 2B = 6 yields B = 3, so y(x) = 3 sin(2x).
Using the chain rule [e^(u(x))]' = u'(x) e^(u(x)) with u(x) = cos(x), we have u'(x) = -sin(x). Therefore, f'(x) = -sin(x) e^(cos(x)).
Rewrite the expression as e^(x ln(1 + 2/x)). As x -> +inf, 2/x -> 0. Using the first-order approximation ln(1 + u) ~ u as u -> 0, we have x ln(1 + 2/x) ~ x * (2/x) = 2. By continuity of the exponential function, the limit is e^2.
Using the quotient rule (u/v)' = (u'v - uv') / v^2 with u = 2x + 1 and v = x + 3: u' = 2 and v' = 1. Thus f'(x) = [2(x + 3) - (2x + 1)(1)] / (x + 3)^2 = [2x + 6 - 2x - 1] / (x + 3)^2 = 5 / (x + 3)^2.
An antiderivative of 3x^2 - 4x + 1 is F(x) = x^3 - 2x^2 + x. Evaluating at the boundaries: F(2) = 2^3 - 2(2^2) + 2 = 8 - 8 + 2 = 2. F(1) = 1^3 - 2(1^2) + 1 = 1 - 2 + 1 = 0. Therefore, F(2) - F(1) = 2 - 0 = 2.
The explicit formula for an arithmetic sequence is u_n = u_0 + n*r. Substituting u_0 = 7 and r = -3 gives u_n = 7 - 3n.
The sum of the first n positive odd numbers is n^2. Here there are 20 terms (from k=0 to 19), so S = 20^2 = 400. Alternatively, using the arithmetic series formula: S = 20 * (u_0 + u_19)/2 = 20 * (1 + 39)/2 = 20 * 20 = 400.
The hereditary property (hérédité) proves transmission of truth: under the inductive hypothesis that P(k) holds for an arbitrary integer k >= n_0, one shows that P(k + 1) must also hold.
For a rational sequence in n, the limit as n -> +inf is given by the ratio of the highest-degree terms: lim (3n^2 + 5n)/(2n^2 - 7) = lim (3n^2)/(2n^2) = 3/2.
The Monotone Convergence Theorem asserts that any sequence of real numbers that is monotonically increasing and bounded from above converges to a finite limit (specifically, its supremum). Similarly, any decreasing sequence bounded below converges.
Compute the terms iteratively: F_0 = 0, F_1 = 1, F_2 = 1 + 0 = 1, F_3 = 1 + 1 = 2, F_4 = 2 + 1 = 3, F_5 = 3 + 2 = 5, and F_6 = 5 + 3 = 8.
Since the function f(x) = sqrt(2 + x) is continuous, the limit L must satisfy L = f(L), so L = sqrt(2 + L). Squaring both sides: L^2 = 2 + L, or L^2 - L - 2 = 0. Factoring yields (L - 2)(L + 1) = 0. Since u_n >= 0 for all n, L must be non-negative, so L = 2.
Since lim (n -> +inf) (-1/n) = 0 and lim (n -> +inf) (1/n) = 0, the sequence (u_n) is pinched between two sequences converging to 0. By the Squeeze Theorem (théorème des gendarmes), lim (n -> +inf) u_n = 0.
Plugging n = 4 into the formula gives S_4 = 4 * (4 + 1) * (2 * 4 + 1) / 6 = 4 * 5 * 9 / 6 = 180 / 6 = 30. Direct summation confirms 1^2 + 2^2 + 3^2 + 4^2 = 1 + 4 + 9 + 16 = 30.
Divide the numerator and denominator by the strictly dominant term 3^n: u_n = [(2/3)^n + 1] / [5 - (2/3)^n]. Since 0 < 2/3 < 1, lim (n -> +inf) (2/3)^n = 0. Therefore, the limit is (0 + 1) / (5 - 0) = 1/5.
The fixed point L satisfies L = (1/3)L + 4, which gives (2/3)L = 4, so L = 4 * (3/2) = 6. Since |1/3| < 1, the arithmetic-geometric sequence converges to its unique fixed point, lim (n -> +inf) u_n = 6.
We can write u_n = (1/n) * (2/n) * ... * (n/n). Since each factor k/n <= 1 for k <= n, we have 0 < u_n <= 1/n for all n >= 1. Since lim (n -> +inf) 1/n = 0, by the Squeeze Theorem, lim u_n = 0.
For each of the n terms in the sum, 1 / (n^2 + n) <= 1 / (n^2 + k) <= 1 / (n^2 + 1). Summing over all n terms yields n / (n^2 + n) <= u_n <= n / (n^2 + 1). Both bounding sequences approach 0 as n -> +inf because degree of numerator is 1 while denominator is 2. By the Squeeze Theorem, lim u_n = 0.
Testing small values: for n = 1, 2^1 = 2 >= 1 (holds); for n = 2, 2^2 = 4 >= 4 (holds); for n = 3, 2^3 = 8 < 3^2 = 9 (fails!); for n = 4, 2^4 = 16 >= 16 (holds). By induction, 2^n >= n^2 holds for all n >= 4. Hence the smallest starting integer n_0 from which the property holds universally is 4.
Multiplying and dividing by the conjugate expression (sqrt(n^2 + 3n) + n) gives [(n^2 + 3n) - n^2] / [sqrt(n^2 + 3n) + n] = 3n / [n * (sqrt(1 + 3/n) + 1)]. Simplifying by n yields 3 / [sqrt(1 + 3/n) + 1]. As n -> +inf, this evaluates to 3 / (1 + 1) = 3/2.
Writing out the telescoping terms: S_n = (1 - 1/2) + (1/2 - 1/3) + ... + (1/n - 1/(n + 1)) = 1 - 1/(n + 1). As n -> +inf, 1/(n + 1) -> 0, so lim S_n = 1 - 0 = 1.
According to standard growth hierarchy theorems in the French Terminale curriculum, the exponential function dominates any polynomial power at +infinity. Therefore, lim (x -> +inf) x^n / e^x = 0 for any positive integer n, meaning lim (x -> +inf) (x^2 + 3x)/e^x = 0.
Using logarithm identities: 3 ln(2) = ln(2^3) = ln(8). Then ln(24) - ln(8) = ln(24 / 8) = ln(3). Finally, ln(3) + ln(9) = ln(3 * 9) = ln(27) = 3 ln(3).
We can write ln(1 + 3x) / x = 3 * [ln(1 + 3x) / (3x)]. As x -> 0, setting u = 3x -> 0 gives the standard limit lim (u -> 0) ln(1 + u)/u = 1. Multiplying by 3 yields 3.
Performing polynomial division: (2x^2 - 5x + 1) = (x - 2)(2x - 1) - 1. Therefore, f(x) = 2x - 1 - 1/(x - 2). As x -> +inf, -1/(x - 2) -> 0, proving that y = 2x - 1 is the slant asymptote.
The derivative is P'(x) = 3x^2 + 3 = 3(x^2 + 1) > 0 for all real x. Therefore, P is continuous and strictly increasing on R. With lim (x -> -inf) P(x) = -inf and lim (x -> +inf) P(x) = +inf, by the corollary to the Intermediate Value Theorem (théorème de la bijection), P(x) = 0 has exactly one unique real solution.
Set u(x) = x => u'(x) = 1, and v'(x) = e^x => v(x) = e^x. By the integration by parts formula: int (0 to 1) x e^x dx = [x e^x]_0^1 - int (0 to 1) e^x dx = (1*e^1 - 0) - [e^x]_0^1 = e - (e^1 - e^0) = e - e + 1 = 1.
By the standard growth hierarchy for the natural logarithm at 0+, power terms dominate logarithms. Thus lim (x -> 0+) x^n ln(x) = 0 for any n > 0, yielding 0.
The derivative is f'(x) = 1*e^(-x) - x*e^(-x) = (1 - x)e^(-x). Since e^(-x) > 0, f'(x) = 0 when x = 1. For x < 1, f'(x) > 0 (increasing), and for x > 1, f'(x) < 0 (decreasing). The maximum is achieved at x = 1, with value f(1) = 1 * e^(-1) = 1 / e.
For a geometric sequence with |q| < 1, the sum of the infinite series is S = v_0 / (1 - q). Here v_0 = 5 and q = 1/2, giving S = 5 / (1 - 1/2) = 5 / (1/2) = 10.
Find the fixed point L satisfying L = 2L - 3 => L = 3. Define the auxiliary sequence v_n = u_n - 3. Then v_(n+1) = u_(n+1) - 3 = (2u_n - 3) - 3 = 2(u_n - 3) = 2 v_n. Thus (v_n) is geometric with ratio 2 and v_0 = u_0 - 3 = 5 - 3 = 2. Hence v_n = 2 * 2^n = 2^(n+1). Returning to u_n: u_n = v_n + 3 = 2^(n+1) + 3.
For any geometric sequence q^n, if |q| < 1, then lim (n -> +inf) q^n = 0. Since |-0.8| = 0.8 < 1, (-0.8)^n converges to 0. Multiplying by 4 gives 4 * 0 = 0.
This is an infinite geometric series with first term a = 1 and common ratio q = -1/3. Since |q| < 1, the sum is S = a / (1 - q) = 1 / (1 - (-1/3)) = 1 / (4/3) = 3/4.