Question

$\lim_{x \to 0} \frac{tan2024x - tan2022x - tan2x}{x^3}$

Answer 1

8185056

Answer 2

Let the given limit be $L$. We have $L = \lim_{x \to 0} \frac{\tan(2024x) - \tan(2022x) - \tan(2x)}{x^3}$. This is a $\frac{0}{0}$ form as $x \to 0$. We can use the Taylor series expansion of $\tan u$ around $u=0$. The Taylor series expansion for $\tan u$ up to the $u^3$ term is $\tan u = u + \frac{u^3}{3} + O(u^5)$.

Let $a = 2024$, $b = 2022$, and $c = 2$. The expression is $\frac{\tan(ax) - \tan(bx) - \tan(cx)}{x^3}$. We notice that $a = b+c$, i.e., $2024 = 2022 + 2$. This relationship will simplify the calculation.

Using the Taylor expansion: $\tan(ax) = ax + \frac{(ax)^3}{3} + O((ax)^5) = ax + \frac{a^3 x^3}{3} + O(x^5)$ $\tan(bx) = bx + \frac{(bx)^3}{3} + O((bx)^5) = bx + \frac{b^3 x^3}{3} + O(x^5)$ $\tan(cx) = cx + \frac{(cx)^3}{3} + O((cx)^5) = cx + \frac{c^3 x^3}{3} + O(x^5)$

Substitute these expansions into the numerator: Numerator $= \left(ax + \frac{a^3 x^3}{3} + O(x^5)\right) - \left(bx + \frac{b^3 x^3}{3} + O(x^5)\right) - \left(cx + \frac{c^3 x^3}{3} + O(x^5)\right)$ Numerator $= (ax - bx - cx) + \left(\frac{a^3 x^3}{3} - \frac{b^3 x^3}{3} - \frac{c^3 x^3}{3}\right) + O(x^5)$ Numerator $= (a - b - c)x + \frac{a^3 - b^3 - c^3}{3} x^3 + O(x^5)$

Since $a = b+c$, we have $a - b - c = 0$. So, the term with $x$ vanishes: Numerator $= (0)x + \frac{a^3 - b^3 - c^3}{3} x^3 + O(x^5)$ Numerator $= \frac{a^3 - b^3 - c^3}{3} x^3 + O(x^5)$

Now, divide the numerator by $x^3$: $\frac{\text{Numerator}}{x^3} = \frac{\frac{a^3 - b^3 - c^3}{3} x^3 + O(x^5)}{x^3} = \frac{a^3 - b^3 - c^3}{3} + O(x^2)$

Take the limit as $x \to 0$: $L = \lim_{x \to 0} \left(\frac{a^3 - b^3 - c^3}{3} + O(x^2)\right) = \frac{a^3 - b^3 - c^3}{3}$

We need to calculate $a^3 - b^3 - c^3$ with $a=2024$, $b=2022$, $c=2$. Since $a = b+c$, we can use the identity: if $x+y+z=0$, then $x^3+y^3+z^3 = 3xyz$. Let $x=a$, $y=-b$, $z=-c$. Then $x+y+z = a - b - c = 0$. So, $a^3 + (-b)^3 + (-c)^3 = 3(a)(-b)(-c)$. $a^3 - b^3 - c^3 = 3abc$.

Substitute this into the limit expression: $L = \frac{3abc}{3} = abc$.

Now, substitute the values of $a$, $b$, and $c$: $L = 2024 \times 2022 \times 2$. $L = 2024 \times (2022 \times 2)$ $L = 2024 \times 4044$.

Calculate the product: $2024 \times 4044 = (2000 + 24) \times (4000 + 44)$ $= 2000 \times 4000 + 2000 \times 44 + 24 \times 4000 + 24 \times 44$ $= 8000000 + 88000 + 96000 + 1056$ $= 8000000 + 184000 + 1056$ $= 8184000 + 1056$ $= 8185056$.

Thus, the limit is 8185056.