Limit of ((1+x)^5 − 1 − 5x) over x²
Evaluate the limit: $$ L = \lim_{x \to 0} \frac{(1+x)^5 - (1+5x)}{x^2 + x^5}. $$
Explanation of Method
Direct substitution gives $\frac{0}{0}$, since $(1+x)^5-(1+5x)\to 0$ and $x^2+x^5\to 0$ as $x\to0$. Rather than differentiating, the fastest route is to expand $(1+x)^5$ fully with the binomial theorem: the constant and linear terms cancel exactly against $1+5x$, leaving a numerator that is visibly a multiple of $x^2$ — the same power as the lowest-order term in the denominator.
- Expand $(1+x)^5$ using the binomial theorem.
- Subtract $1+5x$ to see which terms survive in the numerator.
- Factor $x^2$ out of both the numerator and denominator and cancel it.
- Substitute $x=0$ into the simplified expression.
Step 1 · Expand $(1+x)^5$
By the binomial theorem:
$$ (1+x)^5 = 1 + 5x + 10x^2 + 10x^3 + 5x^4 + x^5. $$
Step 2 · Subtract $1+5x$ from the expansion
The numerator becomes:
$$ (1+x)^5 - (1+5x) = \left(1+5x+10x^2+10x^3+5x^4+x^5\right) - (1+5x) = 10x^2+10x^3+5x^4+x^5. $$
Notice the constant term and the linear term $5x$ cancel exactly, leaving the numerator with lowest-order term $10x^2$ — matching the lowest-order term $x^2$ of the denominator.
Step 3 · Factor out $x^2$ and cancel
The full expression is:
$$ \frac{10x^2+10x^3+5x^4+x^5}{x^2+x^5}. $$
Factor $x^2$ from both the numerator and the denominator:
$$ \frac{x^2\left(10+10x+5x^2+x^3\right)}{x^2\left(1+x^3\right)} = \frac{10+10x+5x^2+x^3}{1+x^3} \quad (x \neq 0). $$
Step 4 · Substitute $x=0$
The simplified expression is a rational function that is continuous at $x=0$ (its denominator there is $1+0^3=1\neq0$), so the limit is found by direct substitution:
$$ L = \frac{10+10(0)+5(0)^2+(0)^3}{1+(0)^3} = \frac{10}{1} = 10. $$
Summary Table
| Expression | Simplified form | Value at $x=0$ |
|---|---|---|
| Numerator | $x^2(10+10x+5x^2+x^3)$ | $0$ (removed by factor $x^2$) |
| Denominator | $x^2(1+x^3)$ | $0$ (removed by factor $x^2$) |
| Ratio after cancelling $x^2$ | $\dfrac{10+10x+5x^2+x^3}{1+x^3}$ | $10$ |
Additional Notes
- The fact that the constant and linear terms cancel exactly is not a coincidence: it happens because $1+5x$ is precisely the first two terms of the binomial expansion of $(1+x)^5$, so subtracting it removes the zeroth- and first-order behavior, leaving an expression of order $x^2$.
- This is the same idea behind the general fact that $(1+x)^n - (1+nx) = \binom{n}{2}x^2 + O(x^3)$ for any $n$, so $\lim_{x\to 0}\frac{(1+x)^n-(1+nx)}{x^2} = \binom{n}{2} = \frac{n(n-1)}{2}$. For $n=5$ this gives $\binom{5}{2}=10$, matching the result here.
- Factoring out the lowest common power of $x$ from both numerator and denominator is a reliable alternative to L'Hôpital's Rule whenever both are polynomials (or start as polynomials after expansion).
Result
The value of the limit is: