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I know how to use the limit process to compute the derivatives of polynomials and trig functions, and it's pretty easy to recognize the limit definition of a derivative.

However, I want to know how to evaluate this limit algebraically and not using L'Hopital with respect to $h$.

$$\lim_{h\to 0} \frac{(x+h)\sqrt{x+h}-(x\sqrt{x})}{h}$$

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  • $\begingroup$ Try using the conjugate. $\endgroup$ – Kenny Lau Apr 30 '17 at 4:41
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    $\begingroup$ It must be $$\lim_{h\to 0}\frac{(x+h)\sqrt{x+h}-x\sqrt x}{h}$$ $\endgroup$ – Ángel Mario Gallegos Apr 30 '17 at 4:42
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    $\begingroup$ oops im high lol $\endgroup$ – Saketh Malyala Apr 30 '17 at 4:45
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HINT:

$$\begin{align} \frac{(x+h)^{3/2}-x^{3/2}}{h}&=\frac{x(x+h)^{1/2}-xx^{1/2}+h(x+h)^{1/2}}{h}\\\\ &=\frac{x\,h}{h((x+h)^{1/2}+x^{1/2})}+(x+h)^{1/2} \end{align}$$

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Hint:

\begin{align*} \lim_{h\to 0}\frac{(x+h)\sqrt{x+h}-x\sqrt x}h&=\lim_{h\to 0}\left(\frac{x\sqrt{x+h}-x\sqrt x}h+\frac{h\sqrt{x+h}}h\right)\\[5pt] &=x\lim_{h\to 0}\frac{\sqrt{x+h}-\sqrt x}h+\sqrt x \end{align*} Use rationalization in order to compute the limit.

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$$\lim_{h\to 0}\frac{(x+h)\sqrt{x+h}-x\sqrt x}{h}=\lim_{h\to 0}\frac{x\sqrt{x+h}-x\sqrt x+h\sqrt{x+h}}{h}$$

$$=\lim_{h\to 0}\frac{x \sqrt x(\sqrt{1+\frac hx}-1)+h\sqrt{x+h}}{h}$$

Now use binomial approximation :

$$ (1+x)^n \approx 1+nx ~~;~ x <<1 $$

$$\lim_{h\to 0}\frac{x \sqrt x \left(\sqrt{1+\frac hx}-1 \right)+h\sqrt{x+h}}{h}=\lim_{h\to 0}\frac{x \sqrt x\left(\dfrac{h}{2x}\right)+h\sqrt{x+h}}{h}$$

$$=\lim_{h\to 0}\frac{h \left(\frac{\sqrt{x}}{2}+\sqrt{x+h} \right)}{h}=\frac{\sqrt{x}}{2}+\lim_{h\to 0}\sqrt{x+h}=\frac{3 \sqrt x}{2}$$

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