Question

If $f(x) = \begin{vmatrix} (x-3)e^x & x & \sin x \\ (x-2)e^x & 1 & \cos x \\ (x-1)e^x & 0 & -\sin x \end{vmatrix}$ then the value of $Lt_{x\to 0} |\frac{f(x)}{x}|$ is equal to

Answer 1

3

Answer 2

To find the value of $Lt_{x\to 0} |\frac{f(x)}{x}|$, we first need to evaluate the determinant $f(x)$.

Given $f(x) = \begin{vmatrix} (x-3)e^x & x & \sin x \\ (x-2)e^x & 1 & \cos x \\ (x-1)e^x & 0 & -\sin x \end{vmatrix}$

Step 1: Evaluate the determinant $f(x)$. We can factor out $e^x$ from the first column ($C_1$): $f(x) = e^x \begin{vmatrix} x-3 & x & \sin x \\ x-2 & 1 & \cos x \\ x-1 & 0 & -\sin x \end{vmatrix}$

Now, let's expand the determinant along the second column ($C_2$) because it contains a zero, which simplifies the calculation: $f(x) = e^x \left[ -x \begin{vmatrix} x-2 & \cos x \\ x-1 & -\sin x \end{vmatrix} + 1 \begin{vmatrix} x-3 & \sin x \\ x-1 & -\sin x \end{vmatrix} - 0 \begin{vmatrix} x-3 & \sin x \\ x-2 & \cos x \end{vmatrix} \right]$

Calculate the $2 \times 2$ determinants: $\begin{vmatrix} x-2 & \cos x \\ x-1 & -\sin x \end{vmatrix} = (x-2)(-\sin x) - (x-1)\cos x = -(x-2)\sin x - (x-1)\cos x$ $\begin{vmatrix} x-3 & \sin x \\ x-1 & -\sin x \end{vmatrix} = (x-3)(-\sin x) - (x-1)\sin x = -(x-3)\sin x - (x-1)\sin x$

Substitute these back into the expression for $f(x)$: $f(x) = e^x \left[ -x (-(x-2)\sin x - (x-1)\cos x) + (-(x-3)\sin x - (x-1)\sin x) \right]$ $f(x) = e^x \left[ x(x-2)\sin x + x(x-1)\cos x - (x-3+x-1)\sin x \right]$ $f(x) = e^x \left[ (x^2-2x)\sin x + (x^2-x)\cos x - (2x-4)\sin x \right]$ Combine the terms with $\sin x$: $f(x) = e^x \left[ (x^2-2x - 2x + 4)\sin x + (x^2-x)\cos x \right]$ $f(x) = e^x \left[ (x^2-4x+4)\sin x + (x^2-x)\cos x \right]$ Recognize the perfect square: $(x^2-4x+4) = (x-2)^2$. $f(x) = e^x \left[ (x-2)^2 \sin x + x(x-1)\cos x \right]$

Step 2: Simplify the expression $\frac{f(x)}{x}$. $\frac{f(x)}{x} = \frac{e^x \left[ (x-2)^2 \sin x + x(x-1)\cos x \right]}{x}$ Divide each term in the bracket by $x$: $\frac{f(x)}{x} = e^x \left[ \frac{(x-2)^2 \sin x}{x} + \frac{x(x-1)\cos x}{x} \right]$ $\frac{f(x)}{x} = e^x \left[ (x-2)^2 \frac{\sin x}{x} + (x-1)\cos x \right]$

Step 3: Evaluate the limit $Lt_{x\to 0} \frac{f(x)}{x}$. Now, we apply the limit $x \to 0$: $Lt_{x\to 0} \frac{f(x)}{x} = Lt_{x\to 0} e^x \left[ (x-2)^2 \frac{\sin x}{x} + (x-1)\cos x \right]$

Using the properties of limits and standard limits:

• $Lt_{x\to 0} e^x = e^0 = 1$
• $Lt_{x\to 0} (x-2)^2 = (0-2)^2 = 4$
• $Lt_{x\to 0} \frac{\sin x}{x} = 1$ (Standard limit)
• $Lt_{x\to 0} (x-1) = (0-1) = -1$
• $Lt_{x\to 0} \cos x = \cos 0 = 1$

Substitute these values into the limit expression: $Lt_{x\to 0} \frac{f(x)}{x} = 1 \left[ (4)(1) + (-1)(1) \right]$ $Lt_{x\to 0} \frac{f(x)}{x} = 1 \left[ 4 - 1 \right]$ $Lt_{x\to 0} \frac{f(x)}{x} = 3$

Step 4: Take the absolute value. The question asks for $Lt_{x\to 0} |\frac{f(x)}{x}|$. $Lt_{x\to 0} |\frac{f(x)}{x}| = |3| = 3$.

The final answer is $\boxed{3}$.