The correct option is c/(c+1).

Analysis of the Given Problem:
Based on the provided question images, we are required to evaluate the limit:
$$\underset{x\to 1}{lim}\left(\frac{1-e^{-c(1-x)}}{1-x{e}^{-c(1-x)}}\right)$$

Step 1: Check the form of the limit
As shown in the second image, if we directly substitute $x=1$ into the expression:
• The numerator becomes:
$$1-e^{-c(1-1)}=1-e^0=1-1=0$$
• The denominator becomes:
$$1-\left(1\right)e^{-c(1-1)}=1-e^0=1-1=0$$
Since the limit yields an indeterminate form of $\frac{0}{0}$, we can apply L'Hôpital's rule.

Step 2: Differentiate the numerator and the denominator
L'Hôpital's rule states that for an indeterminate form of $\frac{0}{0}$, the limit is equal to the limit of the ratio of their derivatives.
Let us find the derivative of the numerator with respect to $x$:
$$\frac{d}{dx}(1-e^{-c(1-x)})=0-e^{-c(1-x)}\cdot \frac{d}{dx}(-c(1-x\left)\right)$$
Since $-c(1-x)=-c+cx$, its derivative with respect to $x$ is $c$. Therefore:
$$\frac{d}{dx}(1-e^{-c(1-x)})=-c{e}^{-c(1-x)}$$

Now, let us differentiate the denominator with respect to $x$ using the product rule:
$$\frac{d}{dx}(1-x{e}^{-c(1-x)})=0-[1\cdot e^{-c(1-x)}+x\cdot \left(c{e}^{-c(1-x)}\right)]$$
Simplifying this expression:
$$\frac{d}{dx}(1-x{e}^{-c(1-x)})=-e^{-c(1-x)}-cx{e}^{-c(1-x)}$$

Step 3: Set up the ratio of derivatives
Substituting the derivatives back into the limit, we get:
$$\underset{x\to 1}{lim}\frac{-c{e}^{-c(1-x)}}{-e^{-c(1-x)}-cx{e}^{-c(1-x)}}$$
We can factor out and cancel the term $-e^{-c(1-x)}$ from both the numerator and the denominator:
$$\underset{x\to 1}{lim}\frac{c}{1+cx}$$

Step 4: Evaluate the limit
Now, substitute $x=1$ into the simplified expression:
$$\frac{c}{1+c\left(1\right)}=\frac{c}{c+1}$$

Thus, the final evaluated limit matches the correct option.