How to find the limit $\displaystyle\lim_{x\rightarrow\infty}\dfrac{(1+x)^n-3}{x^n}?$

Question

I want to find the limit of the $\displaystyle\lim_{x\rightarrow\infty}\dfrac{(1+x)^n-3}{x^n}?$

I know when $n=1,$ then the limiting value is $1,$ because in this case, $\displaystyle\lim_{x\rightarrow\infty}\dfrac{(1+x)^n-3}{x^n}=\displaystyle\lim_{x\rightarrow\infty}\bigg[1-\dfrac{2}{x}\bigg]=2$

Similarly, if $n=2,...$ we get $\displaystyle\lim_{x\rightarrow\infty}\dfrac{(1+x)^n-3}{x^n}=1$ But I do not know, how to generalize the result for general $n$ and discuss the limit.

Answer 1

A good first approach to this type of question is to divide top and bottom by the highest power of $x$ in the expression. So if we divide by $x^n$ we get

$$\lim_{x\to \infty} \frac{(x+1)^n-3}{x^n} = \lim_{x\to \infty} \frac{(x^n + nx^{n-1}+\ldots+1)-3}{x^n} = \lim_{x\to \infty} \frac{(1 + \frac{n}{x}+\ldots+\frac{1}{x^n})-\frac{3}{x^n}}{1}. $$

Can you take it from here?

Answer 2

We simply have that

$$\dfrac{(1+x)^n-3}{x^n}=\dfrac{(1+x)^n}{x^n}-\dfrac{3}{x^n}=\left(1+\dfrac{1}{x}\right)^n-\dfrac{3}{x^n} \to (1+0)^n-0=1^n-0=1 $$