Animated Solution for Mathematics - Limits, Continuity and Differentiability: ${\lim }_{x\to 0}\frac{e-(1+2x{)}^{\frac{1}{2x}}}{x}$ is equal to

Conquering the Exponential Beast

Welcome, fellow traveler on the JEE journey. Today, we are going to tackle a problem that often makes students freeze: the limit of an expression where the variable $x$ is trapped in both the base and the exponent.

We are looking at:

It looks intimidating, doesn't it? But remember, in mathematics, intimidation is just a sign that we need to change our perspective.

The Indeterminate Form

Before we do anything, we must respect the ritual of limits. We plug in $x=0$.

The denominator is clearly $0$. The numerator? We have $e-(1+0{)}^{\frac{1}{0}}$.

As $x\to 0$, $(1+2x{)}^{\frac{1}{2x}}$ approaches the fundamental limit $e$. So, we get $e-e=0$. We are staring at a $\frac{0}{0}$ indeterminate form. This is our green light to proceed.

The Logarithmic Key

How do we handle $(1+2x{)}^{\frac{1}{2x}}$? The variable in the exponent is the enemy of standard differentiation. We need to bring it down.

We use the most powerful identity in our toolkit: $a^b=e^{b\ln a}$. By applying this, our term transforms into $e^{\frac{1}{2x}\ln (1+2x)}$.

Now, the expression looks like:

This is much better. We have moved the complexity into the exponent, where it is easier to manage.

Forcing the Standard Limit

We want to use the standard limit ${\lim }_{t\to 0}\frac{e^t-1}{t}=1$. But our numerator is $e-e^{f(x)}$.

To fix this, we perform a surgical strike: we factor out a $-e$. Now the numerator becomes:

Look at that! We have created the $e^t-1$ structure, where $t=\frac{\ln (1+2x)}{2x}-1$. As $x\to 0$, $t$ also approaches $0$. We are perfectly set up.

The Taylor Series Surgical Strike

After the exponential part simplifies to $1$, we are left with the limit of the exponent itself:

Taking a common denominator, we get:

Now, let's use the Taylor series expansion for $\ln (1+u)=u-\frac{u^2}{2}+\dots$. Substituting $u=2x$, we get $\ln (1+2x)\approx 2x-\frac{(2x{)}^2}{2}=2x-2x^2$.

When we subtract $2x$, the linear terms vanish! We are left with $-2x^2$ in the numerator. The $x^2$ terms cancel out, leaving us with $-1$.

Multiplying by our factored $-e$, we arrive at the beautiful, elegant answer: $e$.

See? The problem wasn't a monster; it was just a puzzle waiting for the right tools. Whenever you see variables in exponents, reach for the logarithmic identity.

Whenever you see standard limits, don't be afraid to manipulate the expression to force them into existence. And when the algebra gets messy, trust the Taylor series to cut through the noise.

Keep practicing, keep questioning, and most importantly, keep falling in love with the elegance of the solution.