Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $m$ and $n$ be two positive integers greater than 1. If ${\lim }_{\alpha \to 0}\left(\frac{e^{\cos ({\alpha }^n)}-e}{{\alpha }^m}\right)=-\left(\frac{e}{2}\right)$ then the value of $m/n$ is .........

Welcome, future engineers! Today, we are performing a surgical operation on the following limit:

At first glance, this looks intimidating. We have an exponential function, a trigonometric function, and unknown powers $m$ and $n$. However, in JEE Advanced, complexity is often just a mask for a very simple, elegant truth waiting to be uncovered.

Algebraic Surgery

Our first step is to simplify the numerator. We observe the expression $e^{\cos ({\alpha }^n)}-e$.

Notice that both terms share a common factor of $e$. Let us factor it out:

We do this because we are hunting for the standard limit ${\lim }_{x\to 0}\frac{e^x-1}{x}=1$. By factoring out $e$, we have isolated the term $e^{\cos ({\alpha }^n)-1}-1$.

If we define $x=\cos ({\alpha }^n)-1$, we see that as $\alpha \to 0$, $x$ also approaches $0$. This serves as our bridge to the standard limit.

The Trigonometric Insight

Now, we must address the term $\cos ({\alpha }^n)-1$. Many students get stuck here by attempting to use L'Hôpital's rule repeatedly, but there is a more efficient path.

Instead, we use the Taylor series expansion. We know that for small $\theta$, $\cos (\theta )\approx 1-\frac{{\theta }^2}{2}$.

Substituting $\theta ={\alpha }^n$, we obtain:

Therefore, $\cos ({\alpha }^n)-1\approx -\frac{{\alpha }^{2n}}{2}$. This approximation is the key that unlocks the entire problem.

The Balancing Act

Now, let us assemble our pieces. Our limit expression becomes:

We can pull the constants out of the limit:

We are given that the final answer is $-\frac{e}{2}$. For this to hold true, the limit of ${\alpha }^{2n-m}$ must be $1$.

The only way a power of $\alpha$ can approach $1$ as $\alpha \to 0$ is if the exponent is zero. Thus, $2n-m=0$, which implies $m=2n$.

Dividing by $n$, we get the final result:

The complexity vanishes, leaving behind a clean, beautiful integer. You have successfully navigated the trap. Keep this mindset: look for the structure, use your tools, and never let the complexity intimidate you.