What is $\lim_{x\to ∞}{(1+x)^k - x^k}\;\;?$

Question

$\lim_{x\to ∞}{(1+x)^k - x^k}$

Can anyone help me with how do you calculate this? If there's no real info about the k parameter?

Also, what are the curly parentheses for? I never seen them before.

Answer 1

The curled braces indicate that you are taking the limit of the whole expression within. You can just as easily use parentheses if you feel like it. Since there is no info about $k$ there is not really anything we can do. We need to know at least that $k$ is a real number and in particular it'd be nice if it was a positive integer or something like that. If it is a positive integer greater than $1$ then the limit will be $\infty$ since we have (using the binomial theorem),

$$(1+x)^k = 1^k+\ldots+x^k$$

Thus the $-x^k$ will be cancelled out and the remaining terms are positive and grow to infinity.

If $k=1$ then we will just have $\lim\limits_{x\to \infty} 1=1$. The $k=0$ case is also obvious.

Answer 2

Tony has covered the case $k \ge 1$. Note for $k < 0$ we can use

$$\frac 1 {(1+x)^k} - \frac 1 {x^k} = \frac {x^k-(1+x)^k}{x^k(1+x)^k}$$

The degree of the numerator is $k-1$, and the degree of the denominator is $2k$, so the limit of $x\to\infty$ is zero.

Answer 3

$\lim_{x\to ∞}{(1+x)^k - x^k} $

$(1+x)^k - x^k =x^k((1+1/x)^k-1) $.

If $f(z) = (1+z)^k$, $f'(0) =\lim_{h \to 0} \dfrac{f(h)-f(0)}{h} =\lim_{h \to 0} \dfrac{(1+h)^k-1}{h} $. But $f'(z) =k(1+z)^{k-1} $ so $f'(0) = k$.

Therefore $\lim_{h \to 0} \dfrac{(1+h)^k-1}{h} =k$ so that $(1+h)^k = 1+hk+o(h) $.

Therefore $(1+x)^k - x^k =x^k((1+1/x)^k-1) =x^k(k/x + o(1/x)) =kx^{k-1} + o(x^{k-1}) $ or $\dfrac{(1+x)^k - x^k}{x^{k-1}} =k + o(1) $.