Let $$f: R \rightarrow R$$ be a differentiable function satisfying $$f'(3) + f'(2) = 0$$. Then $$\lim_{x \to 0} \frac{1 + f(3 + x) - f(3)}{1 + f(2 - x) - f(2)}^{\frac{1}{x}}$$ is equal to:

We have a differentiable function $$f:\; \mathbb R \to \mathbb R$$ which satisfies $$f'(3)+f'(2)=0.$$

We are asked to evaluate

$$\displaystyle \lim_{x\to 0}\Bigg(\frac{1+f(3+x)-f(3)}{1+f(2-x)-f(2)}\Bigg)^{\frac1x}.$$

Because the limit concerns a quotient whose terms approach $$1$$, it is natural to expand each term in the numerator and the denominator by the first-order Taylor formula.

Formula stated: For a differentiable function $$g$$ at a point $$a$$,

$$g(a+h)=g(a)+g'(a)\,h+o(h)\quad\text{as }h\to 0.$$

Applying this to $$f(3+x)$$ about $$x=0$$, we get

$$f(3+x)=f(3)+f'(3)\,x+o(x).$$

Substituting into the expression in the numerator,

$$1+f(3+x)-f(3)=1+\bigl(f(3)+f'(3)x+o(x)\bigr)-f(3)=1+f'(3)\,x+o(x).$$

Next, apply the same Taylor expansion to $$f(2-x)$$ about $$x=0$$:

$$f(2-x)=f(2)+f'(2)\,(-x)+o(x)=f(2)-f'(2)\,x+o(x).$$

So the denominator becomes

$$1+f(2-x)-f(2)=1+\bigl(f(2)-f'(2)x+o(x)\bigr)-f(2)=1-f'(2)\,x+o(x).$$

Because the given information tells us $$f'(3)+f'(2)=0$$, we can set

$$a=f'(3)=-f'(2).$$

Now rewrite both factors with this common constant $$a$$:

Numerator: $$1+f'(3)\,x+o(x)=1+a\,x+o(x).$$

Denominator: $$1-f'(2)\,x+o(x)=1+a\,x+o(x).$$

Hence the entire fraction is

$$\frac{1+a\,x+o(x)}{1+a\,x+o(x)}.$$

To simplify, recall that for small $$u$$,

$$\frac1{1+u}=1-u+o(u).$$

Using this,

$$\frac{1+a\,x+o(x)}{1+a\,x+o(x)}=(1+a\,x+o(x))\bigl(1-a\,x+o(x)\bigr)=1+o(x).$$

Thus we have shown that

$$\frac{1+f(3+x)-f(3)}{1+f(2-x)-f(2)}=1+o(x)\quad\text{as }x\to 0.$$

Let us call this quotient $$1+\phi(x)$$, where $$\phi(x)=o(x)$$, i.e. $$\dfrac{\phi(x)}{x}\to 0$$ when $$x\to 0$$.

Our limit becomes

$$\lim_{x\to 0}\bigl(1+\phi(x)\bigr)^{\frac1x}.$$

Take natural logarithms to convert the exponential form into a product:

$$\ln\!\Bigl[\bigl(1+\phi(x)\bigr)^{\frac1x}\Bigr]=\frac1x\,\ln\bigl(1+\phi(x)\bigr).$$

Again, for small $$u,$$ the logarithmic expansion is $$\ln(1+u)=u-\dfrac{u^{2}}{2}+o(u^{2}).$$ Setting $$u=\phi(x),$$ we have

$$\ln\bigl(1+\phi(x)\bigr)=\phi(x)+o(\phi(x)),$$

and therefore

$$\frac1x\,\ln\bigl(1+\phi(x)\bigr)=\frac{\phi(x)}{x}+o\!\Bigl(\frac{\phi(x)}{x}\Bigr).$$

But by construction $$\dfrac{\phi(x)}{x}\to 0,$$ so the entire right-hand side tends to $$0$$.

Consequently,

$$\lim_{x\to 0}\ln\!\Bigl[\bigl(1+\phi(x)\bigr)^{\frac1x}\Bigr]=0,$$

and exponentiating both sides gives

$$\lim_{x\to 0}\Bigl(1+\phi(x)\Bigr)^{\frac1x}=e^{\,0}=1.$$

Since this limit is exactly the one required, we conclude that

$$\lim_{x\to 0}\Bigg(\frac{1+f(3+x)-f(3)}{1+f(2-x)-f(2)}\Bigg)^{\frac1x}=1.$$

Hence, the correct answer is Option A.