$\lim_{x o 0}\dfrac{(x + 1)^5 - 1}{x}$

Question

$\lim_{x	o 0}\dfrac{(x + 1)^5 - 1}{x}$

Accepted Answer

> [!TLDR]
> Recognize the limit as the derivative of $(x+1)^5$ at $x=0$, or expand the binomial and cancel $x$; either way the polynomial's linear term dominates near zero, giving **5**.

When we see a limit of the form $\frac{f(a+h)-f(a)}{h}$ as $h 	o 0$, we're looking at the definition of the derivative $f'(a)$. Here, if we set $f(t) = t^5$ and shift our variable, the numerator $(x+1)^5 - 1$ is exactly $f(x+1) - f(1)$, which hints at a derivative structure.

But there's a more elementary path that reveals *why* this works: near $x=0$, the polynomial $(x+1)^5$ behaves like $1 + 5x$ plus higher-order terms that vanish faster than $x$ itself. The limit isolates the coefficient of that linear term.

Let me show both perspectives.

## Method 1: Binomial expansion

1. **Expand $(x+1)^5$ using the binomial theorem:**
   $$
   (x+1)^5 = \binom{5}{0}x^0 + \binom{5}{1}x^1 + \binom{5}{2}x^2 + \binom{5}{3}x^3 + \binom{5}{4}x^4 + \binom{5}{5}x^5
   $$
   $$
   = 1 + 5x + 10x^2 + 10x^3 + 5x^4 + x^5
   $$

2. **Substitute into the numerator:**
   $$
   (x+1)^5 - 1 = 5x + 10x^2 + 10x^3 + 5x^4 + x^5
   $$

3. **Divide every term by $x$ (valid since $x 
eq 0$ in the limit process):**
   $$
   \frac{(x+1)^5 - 1}{x} = 5 + 10x + 10x^2 + 5x^3 + x^4
   $$

4. **Take the limit as $x 	o 0$:**
   Each power of $x$ vanishes, leaving only the constant term:
   $$
   \lim_{x 	o 0} \left(5 + 10x + 10x^2 + 5x^3 + x^4ight) = 5
   $$

## Method 2: Derivative interpretation …

Q. $\lim_{x\to 0}\dfrac{(x + 1)^5 - 1}{x}$

Concept understanding — Standard Limit of (xⁿ − aⁿ)/(x − a)

Method 1: Binomial expansion

Method 2: Derivative interpretation …

Q.lim⁡x→0(x+1)5−1x\lim_{x\to 0}\dfrac{(x + 1)^5 - 1}{x}limx→0​x(x+1)5−1​

Method 1: Binomial expansion

Method 2: Derivative interpretation …

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Q. $\lim_{x\to 0}\dfrac{(x + 1)^5 - 1}{x}$

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