Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $\beta ={\lim }_{x\to 0}\frac{\alpha x-(e^{3x}-1)}{\alpha x(e^{3x}-1)}$ for some $\alpha \in \mathbb{R}$. Then the value of $\alpha +\beta$ is :

We are tasked with evaluating the limit expression:

At first glance, this is an indeterminate form of type $\frac{0}{0}$. We must determine the values of $\alpha$ and $\beta$ such that the limit exists and is finite.

To understand the behavior of the denominator, we utilize the standard limit ${\lim }_{x\to 0}\frac{e^{3x}-1}{3x}=1$. We can rewrite the denominator as follows:

As $x\to 0$, the term in the parentheses approaches $1$. Thus, the denominator behaves asymptotically like $3\alpha x^2$. This indicates that the denominator is an $O(x^2)$ object.

We now examine the numerator: $\alpha x-(e^{3x}-1)$. We invoke the Taylor expansion for the exponential function, $e^z=1+z+\frac{z^2}{2!}+\dots$, substituting $z=3x$:

Substituting this into the numerator, the constant terms cancel out:

We now have a numerator of the form $(\alpha -3)x-\frac{9x^2}{2}$ and a denominator behaving like $3\alpha x^2$. If the coefficient $(\alpha -3)$ were non-zero, the limit would be dominated by a term proportional to $\frac{1}{x}$, which diverges as $x\to 0$.

Since $\beta$ is given as a finite value, we must force the coefficient of $x$ to be zero:

With $\alpha =3$ established, the numerator simplifies to $-\frac{9x^2}{2}$. Substituting this back into the limit expression:

The $x^2$ terms cancel out, yielding:

We have successfully determined $\alpha =3$ and $\beta =-1/2$. The final result is: