Vrftsjtryk

$$lim_{h rightarrow 0} {(x+h)^{99}-x^{99}over h}$$

I need to factor this in order to get a limit.

I tried:

$$lim_{h rightarrow 0} {[(x+h)^{33}]^3-[(x)^{33}]^3over h}
\ lim_{h rightarrow 0} {[(x+h)^{33}-(x)^{33}][(x+h)^{66}+(x+h)^{33}(x)^{33}+(x)^{66}]over h}$$
However this does not seem helpful.

How do I approach this question?

If you're familiar with derivatives, then you can recall that
$$ f'(x) = lim_{hto 0} frac{f(x+h)-f(x)}{h} $$
and consider $f(x) = x^{99}$.

Otherwise, use the binomial theorem:
begin{align}frac{(x+h)^{99}-x^{99}}{h} &= frac{1}{h}left( sum_{k=0}^{99} binom{99}{k}x^{99-k}h^{k} - x^{99} right) = frac{1}{h}sum_{k=1}^{99}binom{99}{k}x^{99-k}h^{k}\&= binom{99}{1}x^{98} + sum_{k=2}^{99}binom{99}{k}x^{99-k}h^{k-1}
to 99x^{98}, quad text{as }hto 0.
end{align}

Hint: That’s basically the derivative of $f(x) = x^{99}$.

$$lim_{h rightarrow 0} {(x+h)^{99}-x^{99}over h}$$

Recall that by Binomial Expansion,

$$(a+b)^n = sum_{r = 0}^{n} {n choose r}a^{n-r}b^r$$

So you get

$$lim_{h rightarrow 0} {color{blue}{{99 choose 0}x^{99}}+{99 choose 1}x^{98}h+{99 choose 2}x^{97}h^2+…+{99 choose n}h^{99}color{red}{-x^{99}}over h}$$

Cancel out the first and last terms. From here, notice if anything can be factored. Also, as another hint (for later simplifications), $${n choose r} = frac{n!}{r!{(n-r)!}}$$

The trick is to realise that $h to 0$ means $h$ is arbitrarily small so we can assume $h < 1$ and so $h > h^2 > h^3$ and that the "higher values of $h$" become "negligible"

So you are thinking way too hard. Just use the binomial theorem to get $(x + h)^{99} = x^{99} + 99hx^{98} + sum_{k=2}^{99} {99 choose k} h^kx^{99-k}$ and hope that must of those higher values will become negligible.

And indeed:

$require{cancel}limlimits_{hto 0}frac {(x + h)^{99}-x^{99}}{h }=$

$limlimits_{hto 0}frac {x^{99} + 99hx^{98} + (sum_{k=2}^{99} {99 choose k} h^kx^{99-k})-x^{99}}{h}=$

$limlimits_{hto 0}frac {color{green}{cancel{x^{99}}} + 99color{red}{cancel h}x^{98} + (sum_{k=2}^{99} {99 choose k} h^{color{red}{cancel {k}}k-1}x^{99-k}) - color{green}{cancel{x^{99}}}}{color{red}{cancel h}}=$

$limlimits_{hto 0}(99x^{98} + sum_{k=2}^{99} {99 choose k}h^{k-1}x^{99-k})=$

$99x^{98} + sum_{k=2}^{99}{99 choose k}(limlimits_{hto 0} h^{k-1})x^{99-k}=$

$99x^{98} + sum_{k=2}^{99}{99 choose k}(0)x^{99-k}=$

$99x^{98} + sum_{k=2}^{99}0=$

$99x^{98}$

======

Doug M had an intriguing hint in the comments:

Considering $a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b+ .... + ab^{n-2}+ b^{99})$ we get:

$(x + h)^{99} - x^{99}=$

$[(x+h) - x][(x+h)^{98} + (x+h)^{97}x + ... + (x+h)x^{97} + x^{98}]=$

$h[(x+h)^{98} + (x+h)^{97}x + ... + (x+h)x^{97} + x^{98}]$

And

$[(x+h)^{98} + (x+h)^{97}x + ... + (x+h)x^{97} + x^{98}]=$

$(x^{98} + text{a bunch of stuff with h as a factor})+(x^{97}cdot x + x(text{a bunch of stuff with h as a factor}))+... (xcdot x^{97} + x^{97}(text{a bunch of stuff with h as a factor})) + x^{98})=$

$99x^{98} + text { a REALLY big bunch of stuff with h as a factor}$

So $frac {(x+h)^{98} - x^{99}}h=$

$frac {h(99x^{98}+ text { a REALLY big bunch of stuff with h as a factor}}h = $

$99x^{98} + text { a REALLY big bunch of stuff with h as a factor}$

... and "$ text { a REALLY big bunch of stuff with h as a factor}$" goes to $0$ as .... it's a REALLY big bunch of stuff with $h$ as a factor.

Okay, .... it was an intriguing idea and I'm glad I did it but.... I don't think it's a very practical way to do it.

Because both $limlimits_{hto 0}((x-h)^{99}-x^{99})=0$ and $limlimits_{hto 0}h=0$ you may also use L'Hospital's rule:

$displaystylelimlimits_{hto 0}frac{(x-h)^{99}-x^{99}}{h}=displaystylelimlimits_{hto 0}frac{frac{d}{dh}((x-h)^{99}-x^{99})}{frac{d}{dh}h}=displaystylelimlimits_{hto 0}frac{99(x-h)^{98}-0}{1}=99x^{98}$

This limit is the derivative of $f(x)=x^{99}$. To get a generalization you can replace $99$ with $n$ where $n in mathbb{R}$. The limit thus translates to $displaystyle lim_{h to 0} frac{(x+h)^{n}-x^n}{h}$.

Using Binomial Theorem, which states that $(a+b)^n=displaystyle sum_{k=0}^n {n choose k} a^{n-k} b^k$. Substituting $begin{pmatrix}a \ bend{pmatrix}=begin{pmatrix}x \ hend{pmatrix}$. $$(x+h)^n=x^n+{n choose 1}x^{n-1} h + {n choose 2} x^{n-2} h^2+ cdots +{n choose n-1}xh^{n-1}+h^n$$ Putting back this result in the expression for the limit. We get the following limit $$displaystyle lim_{h to 0} frac{(x+h)^n-x^n}{h}=displaystyle lim_{h to 0} Biggl(frac{1}{h}{n choose 1}x^{n-1}h+frac{1}{h}biggl(h^2biggl({n choose 2}x^{n-2}+{n choose 3}x^{n-3}h+cdots+h^{n-2}biggr)biggr)Biggr)$$ $$=displaystyle lim_{h to 0} Biggl(biggl(nx^{n-1}biggr)+hbiggl({n choose 2}x^{n-2}+{n choose 3}x^{n-3}h+cdots+h^{n-2}biggr)Biggr)$$ $$=displaystyle lim_{h to 0}nx^{n-1}+underbrace{displaystyle lim_{h to 0}hbiggl({n choose 2}x^{n-2}+{n choose 3}x^{n-3}h+cdots+h^{n-2}biggr)}_{= 0}$$
$$implies displaystyle lim_{h to 0}frac{(x+h)^n-x^n}{h}= nx^{n-1}$$
Therefore, For $n=99$, the given limit evaluates to $99x^{98}$. Cheers