Animated Solution for Mathematics - Limits, Continuity and Differentiability: ${\lim }_{x\to \infty }{\left(\frac{x+6}{x+1}\right)}^{x+4}=.........$

We are tasked with evaluating the following limit:

At first glance, this might seem intimidating. However, in the world of JEE Advanced, intimidation is often just a sign that you are about to learn something profound.

Identifying the $1^{\infty }$ Form

The very first rule of the limit club is to always check the indeterminate form before rushing into algebra.

Look at the base: as $x$ becomes extremely large, the ratio $\frac{x+6}{x+1}$ approaches $1$.

Now, look at the exponent: $x+4$ clearly shoots off to infinity. We have a classic $1^{\infty }$ indeterminate form.

The Power of $e$

Whenever we encounter a $1^{\infty }$ form, we use a standard, incredibly powerful shortcut. If $f(x)\to 1$ and $g(x)\to \infty$, the limit transforms into $e$ raised to the power of the limit of $[f(x)-1]\cdot g(x)$.

Mathematically, we use the following identity:

This formula is your best friend. It will save you from the tedious process of taking logarithms and using L'Hopital's rule.

Simplifying $f(x)-1$

Let us substitute our values where $f(x)=\frac{x+6}{x+1}$ and $g(x)=x+4$. Our expression moves into the exponent of $e$:

Now, we simplify the term inside the bracket: $\frac{x+6}{x+1}-1$. Taking the common denominator, the numerator becomes $(x+6)-(x+1)$, which simplifies beautifully to $5$.

The bracket reduces to $\frac{5}{x+1}$. Our expression is now:

Evaluating the Limit

Now, we evaluate the limit sitting in the exponent:

When $x$ tends to infinity, we divide the numerator and denominator by the highest power of $x$. We get:

As $x\to \infty$, the terms $\frac{4}{x}$ and $\frac{1}{x}$ both shrink to zero. We are left with $5\cdot \frac{1+0}{1+0}$, which is exactly $5$.

Putting this value back into the power of $e$, our final answer is $e^5$.