1

$$\lim_{h\to 0}\frac{\Bigl((x+h)^{\frac{3}{4}}-(x)^{\frac{3}{4}}\Bigr)}{h}$$

I understand the process till

$$\lim_{h\to 0}\frac{\Bigl((x+h)^{\frac34}-(x)^{\frac{3}{4}}\Bigr)}{h} * \frac{\Bigl((x+h)^{\frac{3}{4}}+(x)^{\frac{3}{4}}\Bigr)}{\Bigl((x+h)^{\frac{3}{4}}+(x)^{\frac{3}{4}}\Bigr)}$$

and post expansion

$$\lim_{h\to 0}\frac{\Bigl(h^3+3h^2x+3x^2h\Bigr)}{{h}\Bigl((x+h)^{\frac{3}{4}}+(x)^{\frac{3}{4}}\Bigr)\Bigl((x+h)^{\frac{3}{2}}+(x)^{\frac{3}{2}}\Bigr)}$$

but beyond this i am unable to reduce to:

$\frac 34\cdot x^{\frac{-1}{4}}$

Feng
  • 13,705
Ashwin Sarith
  • 49
  • 1
    The last step is divide numerator and denominator with $h$ then your function is continuous so you can just replace $h$ with $0$. – kingW3 Sep 09 '19 at 11:54

2 Answers2

3

Why don't you expand from the start

$$\lim_{h\to0}\dfrac{(x+h)^n-x^n}h=x^n\cdot\lim_{h\to0}\dfrac{\left(1+\dfrac hx\right)^n-1}h$$

Now use Binomial series

Alternatively, set $$(x+h)^{1/4}=a,x+h=a^4; x^{1/4}=b, x=b^4$$

$$\lim_{h\to0}\dfrac{(x+h)^{3/4}-x^{1/4}}h=\lim_{a\to b}\dfrac{a^3-b^3}{a^4-b^4}=\lim_{a\to b}\dfrac{a^2+ab+b^2}{a^3+a^2b+ab^2+b^3}=\dfrac{3b^2}{4b^3}=\dfrac3{4b}=\dfrac3{4x^{1/4}}$$

lab bhattacharjee
  • 274,582
1

\begin{align*} \lim_{h\to 0}\frac{\Bigl(h^3+3h^2x+3x^2h\Bigr)}{{h}\Bigl((x+h)^{\frac{3}{4}}+(x)^{\frac{3}{4}}\Bigr)\Bigl((x+h)^{\frac{3}{2}}+(x)^{\frac{3}{2}}\Bigr)}&=\lim_{h\to0}\frac{h^3+3h^2x+3hx^2}{h}\lim_{h\to0}\frac1{(x+h)^{\frac{3}{4}}+(x)^{\frac{3}{4}}}\lim_{h\to0}\frac{1}{(x+h)^{\frac{3}{2}}+(x)^{\frac{3}{2}}}\\&=3x^2\cdot\frac1{2x^{\frac34}}\cdot\frac1{2x^{\frac32}}\\&=\frac34x^{-\frac14}. \end{align*}

Feng
  • 13,705