Let `f (x) = lim _(n to oo) n ^(2) tan (ln(sec""(x)/(n )))and g (x) = min (f(x), {x}}`
(where {.} denotes fractional part function)
Left derivative of `phi(x) =e ^(sqrt(2f (x))) at x =0` is:

Options

A. 0
B. 1
C. `-1`
D. Does not exist

Answer

To solve the problem step by step, we will first analyze the function \( f(x) \) and then find the left derivative of \( \phi(x) \) at \( x = 0 \).

### Step 1: Find \( f(x) \)

Given:
\[ f(x) = \lim_{n \to \infty} n^2 \tan\left(\ln\left(\frac{1}{\cos(x)}\right) / n\right) \]

We can rewrite this limit by substituting \( t = \frac{1}{n} \), which implies \( n = \frac{1}{t} \) and as \( n \to \infty \), \( t \to 0^+ \).

Thus, we have:
\[ f(x) = \lim_{t \to 0^+} \left(\frac{1}{t^2} \tan\left(-\ln(\cos(x)) t\right)\right) \]

### Step 2: Simplify \( f(x) \)

Using the small angle approximation for \( \tan \):
\[ \tan(u) \approx u \text{ as } u \to 0 \]
we can write:
\[ \tan\left(-\ln(\cos(x)) t\right) \approx -\ln(\cos(x)) t \]

Substituting this back into our limit gives:
\[ f(x) = \lim_{t \to 0^+} \left(\frac{1}{t^2} \cdot (-\ln(\cos(x)) t)\right) = \lim_{t \to 0^+} \left(-\frac{\ln(\cos(x))}{t}\right) \]

As \( t \to 0 \), this expression diverges unless \( \ln(\cos(x)) = 0 \), which occurs at \( x = 0 \). Thus, we can conclude:
\[ f(0) = 0 \]

### Step 3: Evaluate \( f(x) \)

For small \( x \):
\[ \cos(x) \approx 1 - \frac{x^2}{2} \]
Thus,
\[ \ln(\cos(x)) \approx -\frac{x^2}{2} \]
This leads to:
\[ f(x) = \lim_{n \to \infty} n^2 \tan\left(-\frac{x^2}{2n}\right) \approx \lim_{n \to \infty} n^2 \left(-\frac{x^2}{2n}\right) = -\frac{x^2}{2} \cdot n \to 0 \]

### Step 4: Find \( \phi(x) \)

Given:
\[ \phi(x) = e^{\sqrt{2f(x)}} \]
Since \( f(0) = 0 \):
\[ \phi(0) = e^{\sqrt{2 \cdot 0}} = e^0 = 1 \]

### Step 5: Left Derivative of \( \phi(x) \) at \( x = 0 \)

To find the left derivative, we need to evaluate:
\[ \phi'(x) = \frac{d}{dx}\left(e^{\sqrt{2f(x)}}\right) \]

Using the chain rule:
\[ \phi'(x) = e^{\sqrt{2f(x)}} \cdot \frac{1}{2\sqrt{2f(x)}} \cdot 2f'(x) \]
At \( x = 0 \):
\[ f'(x) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{f(h)}{h} \]
Since \( f(h) \approx -\frac{h^2}{2} \), we find:
\[ f'(0) = \lim_{h \to 0} \frac{-\frac{h^2}{2}}{h} = \lim_{h \to 0} -\frac{h}{2} = 0 \]

Thus:
\[ \phi'(0) = e^0 \cdot \frac{1}{0} \cdot 0 = 0 \]

However, we need the left-hand derivative:
For \( x < 0 \):
\[ \phi(x) = e^{\sqrt{2f(x)}} \approx e^{\sqrt{-x^2}} = e^{-|x|} \text{ (as \( x \to 0^- \))} \]
Differentiating:
\[ \phi'(x) = -e^{-|x|} \text{ for } x < 0 \]
At \( x = 0 \):
\[ \phi'(0^-) = -1 \]

### Final Answer

The left derivative of \( \phi(x) \) at \( x = 0 \) is:
\[ \boxed{-1} \]