Evaluate: `lim_(xto-1)(log(x^(2)+2x+5)+"cos5"((x+1))/6-"cos"((x+1))/6-log4)/((x+1)^(2))`

Options

A. `-1/6`
B. `-1/12`
C. `-1/4`
D. `-1/8`

Answer

To evaluate the limit

\[ \lim_{x \to -1} \frac{\log(x^2 + 2x + 5) + \cos\left(\frac{5(x+1)}{6}\right) - \cos\left(\frac{x+1}{6}\right) - \log 4}{(x+1)^2} \]

we will follow these steps:

### Step 1: Substitute \( x + 1 = h \)

Let \( h = x + 1 \). As \( x \to -1 \), \( h \to 0 \). Therefore, we can rewrite the limit in terms of \( h \):

\[ \lim_{h \to 0} \frac{\log((h-1)^2 + 2(h-1) + 5) + \cos\left(\frac{5h}{6}\right) - \cos\left(\frac{h}{6}\right) - \log 4}{h^2} \]

### Step 2: Simplify the expression inside the logarithm

Now simplify \( (h-1)^2 + 2(h-1) + 5 \):

\[ (h-1)^2 + 2(h-1) + 5 = h^2 - 2h + 1 + 2h - 2 + 5 = h^2 + 4 \]

So we have:

\[ \lim_{h \to 0} \frac{\log(h^2 + 4) + \cos\left(\frac{5h}{6}\right) - \cos\left(\frac{h}{6}\right) - \log 4}{h^2} \]

### Step 3: Rewrite the limit

This becomes:

\[ \lim_{h \to 0} \frac{\log\left(\frac{h^2 + 4}{4}\right) + \cos\left(\frac{5h}{6}\right) - \cos\left(\frac{h}{6}\right)}{h^2} \]

### Step 4: Evaluate the limit

As \( h \to 0 \):

- \( \log\left(\frac{h^2 + 4}{4}\right) = \log\left(1 + \frac{h^2}{4}\right) \) which approaches \( \frac{h^2}{4} \) (using the approximation \( \log(1+x) \approx x \) for small \( x \)).
- \( \cos\left(\frac{5h}{6}\right) \) approaches \( 1 \) and \( \cos\left(\frac{h}{6}\right) \) also approaches \( 1 \). Thus, \( \cos\left(\frac{5h}{6}\right) - \cos\left(\frac{h}{6}\right) \) approaches \( 0 \).

Using Taylor expansion:

\[ \cos\left(\frac{5h}{6}\right) \approx 1 - \frac{(5h/6)^2}{2} = 1 - \frac{25h^2}{72} \]

\[ \cos\left(\frac{h}{6}\right) \approx 1 - \frac{(h/6)^2}{2} = 1 - \frac{h^2}{72} \]

Thus,

\[ \cos\left(\frac{5h}{6}\right) - \cos\left(\frac{h}{6}\right) \approx -\frac{25h^2}{72} + \frac{h^2}{72} = -\frac{24h^2}{72} = -\frac{h^2}{3} \]

### Step 5: Substitute back into the limit

Now we have:

\[ \lim_{h \to 0} \frac{\frac{h^2}{4} - \frac{h^2}{3}}{h^2} = \lim_{h \to 0} \left(\frac{1}{4} - \frac{1}{3}\right) = \frac{3 - 4}{12} = -\frac{1}{12} \]

### Final Answer

Thus, the limit evaluates to:

\[ \boxed{-\frac{1}{12}} \]