Question

Let $f(x) = 3x^{10}-7x^8+5x^6-21x^3+3x^2-7$

$265 \left( \lim\limits_{h \to 0} \frac{h^4+3h^2}{(f(1-h)-f(1)) \sin 5h} \right) =$

• A. 1
• B. 2
• C. 3
• D. -3

Answer 1

Answer: (C)

3

Answer 2

The given expression is $265 \left( \lim\limits_{h \to 0} \frac{h^4+3h^2}{(f(1-h)-f(1)) \sin 5h} \right)$. Let's evaluate the limit part: $L = \lim\limits_{h \to 0} \frac{h^4+3h^2}{(f(1-h)-f(1)) \sin 5h}$.

As $h \to 0$, the numerator $h^4+3h^2 \to 0^4+3(0)^2 = 0$. As $h \to 0$, $1-h \to 1$, so $f(1-h) \to f(1)$. Thus, the term $f(1-h)-f(1) \to 0$. Also, as $h \to 0$, $\sin 5h \to \sin(0) = 0$. The limit is of the indeterminate form $\frac{0}{0}$.

We can rewrite the expression to use the definition of the derivative $f'(a) = \lim\limits_{x \to a} \frac{f(x)-f(a)}{x-a}$. Let's consider the term $f(1-h)-f(1)$. Let $x = 1-h$. As $h \to 0$, $x \to 1$. Then $f(1-h)-f(1) = f(x)-f(1)$. The denominator difference is $(1-h)-1 = -h$. So, $\lim\limits_{h \to 0} \frac{f(1-h)-f(1)}{-h} = \lim\limits_{x \to 1} \frac{f(x)-f(1)}{x-1} = f'(1)$. Therefore, for small $h$, $f(1-h)-f(1) \approx -h f'(1)$.

We also use the standard limit $\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$. For small $h$, $\sin 5h \approx 5h$.

Let's rewrite the limit expression by dividing the numerator and denominator by appropriate powers of $h$: $L = \lim\limits_{h \to 0} \frac{h^2(h^2+3)}{(f(1-h)-f(1)) \sin 5h}$ $L = \lim\limits_{h \to 0} \frac{h^2(h^2+3)}{\left(\frac{f(1-h)-f(1)}{-h}\right) (-h) \left(\frac{\sin 5h}{5h}\right) (5h)}$ $L = \lim\limits_{h \to 0} \frac{h^2(h^2+3)}{(-5h^2) \left(\frac{f(1-h)-f(1)}{-h}\right) \left(\frac{\sin 5h}{5h}\right)}$ For $h \neq 0$, we can cancel $h^2$ from the numerator and denominator: $L = \lim\limits_{h \to 0} \frac{h^2+3}{-5 \left(\frac{f(1-h)-f(1)}{-h}\right) \left(\frac{\sin 5h}{5h}\right)}$

Now, we evaluate the limit of each part as $h \to 0$: $\lim\limits_{h \to 0} (h^2+3) = 0^2+3 = 3$. $\lim\limits_{h \to 0} \frac{f(1-h)-f(1)}{-h} = f'(1)$ by the definition of the derivative. $\lim\limits_{h \to 0} \frac{\sin 5h}{5h} = 1$ by the standard limit.

So, the limit $L$ is: $L = \frac{3}{-5 \cdot f'(1) \cdot 1} = \frac{3}{-5 f'(1)}$.

Now we need to find $f'(1)$. $f(x) = 3x^{10}-7x^8+5x^6-21x^3+3x^2-7$ The derivative $f'(x)$ is: $f'(x) = \frac{d}{dx}(3x^{10}) - \frac{d}{dx}(7x^8) + \frac{d}{dx}(5x^6) - \frac{d}{dx}(21x^3) + \frac{d}{dx}(3x^2) - \frac{d}{dx}(7)$ $f'(x) = 3(10x^9) - 7(8x^7) + 5(6x^5) - 21(3x^2) + 3(2x) - 0$ $f'(x) = 30x^9 - 56x^7 + 30x^5 - 63x^2 + 6x$.

Now, evaluate $f'(1)$: $f'(1) = 30(1)^9 - 56(1)^7 + 30(1)^5 - 63(1)^2 + 6(1)$ $f'(1) = 30 - 56 + 30 - 63 + 6$ $f'(1) = (30 + 30 + 6) - (56 + 63)$ $f'(1) = 66 - 119$ $f'(1) = -53$.

Substitute the value of $f'(1)$ into the limit expression: $L = \frac{3}{-5(-53)} = \frac{3}{265}$.

The question asks for the value of $265 \times L$. $265 \times L = 265 \times \frac{3}{265} = 3$.

The final answer is 3.