Let u and $$\upsilon$$ be the distances of the object and the image from a lens of focal length $$f$$. The correct graphical representation of u and $$\upsilon$$ for a convex lens when $$|u| > f$$, is

$$\text{Object distance} = -u \quad (u > 0)$$

$$\text{Focal length} = +f$$

$$\frac{1}{v} - \frac{1}{-u} = \frac{1}{f} \implies \frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$

Isolating image distance $$v$$: $$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} \implies \frac{1}{v} = \frac{u - f}{uf} \implies v = \frac{uf}{u - f}$$

Given condition $$|u| > f \implies u > f$$:

$$\text{As } u \to f^{+}, \quad v \to +\infty$$

$$\text{As } u \to +\infty, \quad v \to f^{+}$$