Question

Find the value of $\underset{x\to 0}{lim}\frac{(1+x{)}^5-1}{x}$

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Answer 1

We will solve this by using the expansion of $(1+x{)}^5$.

We saw it in binomial theorem that $(a+b{)}^n$ can be expanded as $(a+b{)}^n=a^n+n{c}_1{a}^{n-1}b+.....b^n$.

We don't need all the terms of $(1+x{)}^5$.

$\Rightarrow (1+x{)}^5=1+5c_{1}x+5c_2{x}^2......x^5$

$\Rightarrow \underset{x\to 0}{lim}\frac{(1+x{)}^5-1}{x}=\underset{x\to 0}{lim}\frac{(1+5c_1.x+5c_2{x}^2....)-1}{x}$

$=\underset{x\to 0}{lim}\frac{5x+5c_2{x}^2....}{x}{x}^5$

$=\underset{x\to 0}{lim}5+5c_{2}x+.....x^4$

$=5$.

We don't need all the terms of $(1+x{)}^5$ because they will become zero once we substitute x=0