Let `f(x)=3x^10-7x^8+5x^6-21x^3+3x^2-7` , then the value of `lim_(h->0) (f(1-h)-f(1))/(h^3+3h)`

Options

A. `50/3`
B. `22/3`
C. `3`
D. `53/3`

Answer

To solve the limit problem, we will follow these steps:

### Step 1: Write down the limit expression
We need to evaluate the limit:
\[ L = \lim_{h \to 0} \frac{f(1-h) - f(1)}{h^3 + 3h} \]

### Step 2: Factor the denominator
Notice that we can factor \(h\) from the denominator:
\[ h^3 + 3h = h(h^2 + 3) \]
Thus, we can rewrite the limit as:
\[ L = \lim_{h \to 0} \frac{f(1-h) - f(1)}{h(h^2 + 3)} \]

### Step 3: Recognize the derivative
The expression \(\frac{f(1-h) - f(1)}{-h}\) resembles the definition of the derivative. We can rewrite the limit as:
\[ L = -\lim_{h \to 0} \frac{f(1-h) - f(1)}{-h(h^2 + 3)} = -\frac{1}{3} \lim_{h \to 0} \frac{f(1-h) - f(1)}{-h} \]
This implies:
\[ L = -\frac{1}{3} f'(1) \]

### Step 4: Calculate \(f'(x)\)
Now we need to find \(f'(x)\). Given:
\[ f(x) = 3x^{10} - 7x^8 + 5x^6 - 21x^3 + 3x^2 - 7 \]
We differentiate \(f(x)\):
\[ f'(x) = 30x^9 - 56x^7 + 30x^5 - 63x^2 + 6x \]

### Step 5: Evaluate \(f'(1)\)
Now we substitute \(x = 1\) into \(f'(x)\):
\[ f'(1) = 30(1)^9 - 56(1)^7 + 30(1)^5 - 63(1)^2 + 6(1) \]
Calculating this gives:
\[ f'(1) = 30 - 56 + 30 - 63 + 6 = -53 \]

### Step 6: Substitute \(f'(1)\) back into the limit
Now we substitute \(f'(1)\) back into our expression for \(L\):
\[ L = -\frac{1}{3}(-53) = \frac{53}{3} \]

### Final Answer
Thus, the value of the limit is:
\[ \boxed{\frac{53}{3}} \]