91Ó°ÊÓ

The given equation of hyperbola $\frac{x^2}{A^2}-\frac{y^2}{B^2}=1$

Where $A,B$are non zero constants

Any conic section can be defined as the locus of points with constant distances to a point and a line. That ratio is known as eccentricity, and it is commonly represented by the symbol $e$

The eccentricity of a hyperbola is defined as

$e=\frac{\sqrt{A^2+B^2}}{A}$

The eccentricity of a hyperbola is greater than $1$

Hence, The foci of a hyperbola are further away from the center and even the vertices, resulting in greater eccentricity.

The eccentricity is given by $e=\frac{\sqrt{A^2+B^2}}{A}$

Then,

$$\begin{gathered} \underset{A→0}{\lim }\frac{\sqrt{A^2+B^2}}{A}=\frac{\sqrt{0^2+B^2}}{0} \\ =∞ \end{gathered}$$

Therefore, the answer is $\underset{A→B}{\lim }\frac{\sqrt{A^2+B^2}}{A}=0$

The hyperbola elongates and flattens.

The eccentricity is given by$e=\frac{\sqrt{A^2+B^2}}{A}.$

Then,

$$\begin{gathered} \underset{A→0}{\lim }e=\underset{A→∞}{\lim }\frac{\sqrt{A^2+B^2}}{A} \\ \underset{A→∞}{\lim }\frac{\sqrt{A^2+B^2}}{A}=\frac{\sqrt{{∞}^2+B^2}}{∞} \\ =\frac{∞}{∞} \\ =1 \end{gathered}$$

Hence the answer is $\underset{A→∞}{\lim }\frac{\sqrt{A^2+B^2}}{A}=1$

The hyperbola becomes more like a parabola.