Let `f(x)=2x^(1//3)+3x^(1//2)+1.` The value of `lim_(hrarr0)(f(1+h)-f(1-h))/(h^(2)+2h)` is equal to

Options

A. 2
B. `(13)/(12)`
C. `(13)/(6)`
D. `(13)/(3)`

Answer

To solve the limit \( \lim_{h \to 0} \frac{f(1+h) - f(1-h)}{h^2 + 2h} \) where \( f(x) = 2x^{1/3} + 3x^{1/2} + 1 \), we will follow these steps:

### Step 1: Evaluate \( f(1+h) \) and \( f(1-h) \)

First, we need to compute \( f(1+h) \) and \( f(1-h) \):

\[ f(1+h) = 2(1+h)^{1/3} + 3(1+h)^{1/2} + 1 \]
\[ f(1-h) = 2(1-h)^{1/3} + 3(1-h)^{1/2} + 1 \]

### Step 2: Substitute into the limit expression

Now, substitute these expressions into the limit:

\[ f(1+h) - f(1-h) = \left(2(1+h)^{1/3} + 3(1+h)^{1/2} + 1\right) - \left(2(1-h)^{1/3} + 3(1-h)^{1/2} + 1\right) \]

This simplifies to:

\[ f(1+h) - f(1-h) = 2\left((1+h)^{1/3} - (1-h)^{1/3}\right) + 3\left((1+h)^{1/2} - (1-h)^{1/2}\right) \]

### Step 3: Apply the limit

Now, we need to compute the limit:

\[ \lim_{h \to 0} \frac{2\left((1+h)^{1/3} - (1-h)^{1/3}\right) + 3\left((1+h)^{1/2} - (1-h)^{1/2}\right)}{h^2 + 2h} \]

### Step 4: Use Taylor expansion for small \( h \)

For small \( h \), we can use the Taylor expansion:

\[ (1+h)^{1/3} \approx 1 + \frac{1}{3}h - \frac{1}{9}h^2 + O(h^3) \]
\[ (1-h)^{1/3} \approx 1 - \frac{1}{3}h - \frac{1}{9}h^2 + O(h^3) \]

Thus,

\[ (1+h)^{1/3} - (1-h)^{1/3} \approx \left(1 + \frac{1}{3}h - \frac{1}{9}h^2\right) - \left(1 - \frac{1}{3}h - \frac{1}{9}h^2\right) = \frac{2}{3}h + O(h^2) \]

And similarly for the square root:

\[ (1+h)^{1/2} \approx 1 + \frac{1}{2}h - \frac{1}{8}h^2 + O(h^3) \]
\[ (1-h)^{1/2} \approx 1 - \frac{1}{2}h - \frac{1}{8}h^2 + O(h^3) \]

Thus,

\[ (1+h)^{1/2} - (1-h)^{1/2} \approx \left(1 + \frac{1}{2}h - \frac{1}{8}h^2\right) - \left(1 - \frac{1}{2}h - \frac{1}{8}h^2\right) = h + O(h^2) \]

### Step 5: Substitute back into the limit

Now substituting back into our limit expression:

\[ f(1+h) - f(1-h) \approx 2\left(\frac{2}{3}h\right) + 3(h) = \frac{4}{3}h + 3h = \left(\frac{4}{3} + 3\right)h = \frac{13}{3}h \]

### Step 6: Simplify the limit

Now we can substitute this into our limit:

\[ \lim_{h \to 0} \frac{\frac{13}{3}h}{h^2 + 2h} = \lim_{h \to 0} \frac{\frac{13}{3}h}{h(h + 2)} = \lim_{h \to 0} \frac{\frac{13}{3}}{h + 2} \]

As \( h \to 0 \), this limit becomes:

\[ \frac{\frac{13}{3}}{2} = \frac{13}{6} \]

### Final Answer

Thus, the value of the limit is:

\[ \frac{13}{6} \]

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