Animated Solution for Mathematics - Limits, Continuity and Differentiability: The integer $n$ for which ${\lim }_{x\to 0}\frac{(\cos x-1)(\cos x-e^x)}{x^n}$ is a finite non-zero number is

Imagine you are standing at the origin of a coordinate plane, looking at the behavior of functions as they approach the point $x=0$. In the world of calculus, this is where the magic happens.

We are presented with a limit:

At first glance, this looks like a standard indeterminate form of type $\frac{0}{0}$. We are tasked with finding an integer $n$ that forces this expression to settle into a finite, non-zero value.

Many students instinctively reach for L'Hopital's rule when they see $\frac{0}{0}$. While valid, it is often a trap.

If you differentiate the product $(\cos x-1)(\cos x-e^x)$, you will find yourself drowning in a sea of product rules and chain rules. Instead, let us use the most powerful tool in our arsenal: the Maclaurin series.

By representing these functions as infinite polynomials, we can see exactly how they behave as $x$ gets infinitesimally close to zero.

Let us break down the numerator into two distinct parts. First, consider $\cos x-1$.

We know the standard expansion for cosine is $\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\dots$. When we subtract $1$, the constant terms vanish, leaving us with:

As $x$ approaches $0$, the term $-\frac{x^2}{2}$ dominates everything else. This is our first leading term.

Next, we tackle the second bracket: $\cos x-e^x$. Using the expansion for $e^x=1+x+\frac{x^2}{2!}+\dots$, we write:

Notice the beauty of the cancellation here. The $1$s cancel out, and we are left with $-x-x^2-\dots$. The leading term here is simply $-x$.

Now, we multiply these two leading terms together to find the behavior of the entire numerator near the origin:

This is the heart of the problem. The numerator behaves like $\frac{x^3}{2}$ when $x$ is very small. Now, look back at our original limit:

To make this limit a finite, non-zero number, the powers of $x$ must vanish. If $n<3$, the $x$ remains in the numerator, and the limit is $0$. If $n>3$, the $x$ remains in the denominator, and the limit shoots off to infinity.

But if $n=3$, the $x^3$ terms cancel out perfectly, leaving us with $L=\frac{1}{2}$.

We have navigated the complexity of transcendental functions by stripping them down to their core power series. By identifying the leading term, we transformed a daunting limit into a simple algebraic comparison.

The integer we sought is $n=3$.

It is a reminder that in mathematics, the most complex-looking problems often yield to the most fundamental principles if we only take the time to look closely at the behavior of the functions involved. Keep exploring, keep calculating, and never fear the limit!