What is the limit of (1+x)^4 - 1 / x as x approaches 0?
Understand the Problem
The question asks to find the limit of the function (1+x)^4 - 1 / x as x approaches 0. This is a calculus problem that involves finding the limit using L'Hopital's rule or by expanding the binomial term and simplifying the expression.
Answer
4
Answer for screen readers
4
Steps to Solve
- Expand $(1+x)^4$ using the binomial theorem
The binomial theorem states that $(a+b)^n = \sum_{k=0}^{n} {n \choose k} a^{n-k} b^k$. In our case, $a=1$, $b=x$, and $n=4$. Thus, $$(1+x)^4 = {4 \choose 0}1^4x^0 + {4 \choose 1}1^3x^1 + {4 \choose 2}1^2x^2 + {4 \choose 3}1^1x^3 + {4 \choose 4}1^0x^4$$ $$(1+x)^4 = 1 + 4x + 6x^2 + 4x^3 + x^4$$
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Substitute the expansion into the limit expression Replacing $(1+x)^4$ in the limit expression, we get: $$ \lim_{x \to 0} \frac{(1 + 4x + 6x^2 + 4x^3 + x^4) - 1}{x} $$
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Simplify the expression
$$ \lim_{x \to 0} \frac{4x + 6x^2 + 4x^3 + x^4}{x} $$ Factor out an $x$ from the numerator: $$ \lim_{x \to 0} \frac{x(4 + 6x + 4x^2 + x^3)}{x} $$ Cancel the $x$ terms: $$ \lim_{x \to 0} (4 + 6x + 4x^2 + x^3) $$
- Evaluate the limit As $x$ approaches $0$, each term with $x$ in it will go to $0$. $$ \lim_{x \to 0} (4 + 6x + 4x^2 + x^3) = 4 + 6(0) + 4(0)^2 + (0)^3 = 4 $$
4
More Information
The limit of the given function as $x$ approaches 0 is 4.
Tips
A common mistake is not correctly expanding the binomial $(1+x)^4$ or making algebraic errors during simplification. Also, some may try L'Hopital's rule but it leads to more complicated derivatives.
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