1. ## [SOLVED] Difficult limit

Lim [(x+h)²ln(x+h) - x²ln(x)] / h
h→0

Of course this is finding the derivative of the obvious function, but I would like to know if anyone can evluate this WITHOUT using L'Hospital.
Thanks.

2. Originally Posted by spammanon
Lim [(x+h)²ln(x+h) - x²ln(x)] / h
h→0

Of course this is finding the derivative of the obvious function, but I would like to know if anyone can evluate this WITHOUT using L'Hospital.
Thanks.
The limit does not make sense unless $\displaystyle x>0$, thus I will asume that $\displaystyle x>0$.

Consider $\displaystyle f(t) = t^2 \ln t$ then then the derivative at $\displaystyle t=x$ is $\displaystyle f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h\to 0} \frac{(x+h)^2 \ln (x+h) - x^2 \ln x}{h}$. This is not using L'Hopital rule this is just the definition of the derivative.

3. ## Thanks, but...

Hello,

As I clearly stated, I know this is the derivative of a function. What I am asking is how to evaluate the limit. That is, rearrange the terms or use other algebraic manipulations so that the limit is clearly seen. So, anyone else?

How do you evaluate the limit of the expression without using L'Hospital's rule.

Thanks.

4. $\displaystyle \lim_{h\to 0} \frac{(x+h)^2 \ln (x+h) - x^2 \ln x}{h} =$$\displaystyle \lim_{h\to 0}\frac{(x+h)^2 \ln (x+h) - (x+h)^2 \ln x + (x+h)^2 \ln x - x^2\ln x}{h}$

$\displaystyle \lim_{h\to 0} ~ (x+h)^2 \cdot \frac{\ln (x+h) - \ln x}{h} + \ln x \cdot \lim_{h\to 0}\frac{(x+h)^2 - x^2}{h}$.

Thus, we get,
$\displaystyle x^2 \lim_{h\to 0}\frac{ \ln (x+h) - \ln x}{h} + 2x\ln x$.

We claim that,
$\displaystyle \lim_{h\to 0}\frac{\ln (x+h) - \ln x}{h} = \frac{1}{x}$.

Note that,
$\displaystyle \ln (x+h) = \int_1^{x+h} \frac{dt}{t} \mbox{ and }\ln x = \int_1^x \frac{dt}{t}$.

Now,
[tex]\frac{1}{x} = \int_x^{x+h} \frac

Thus, $\displaystyle \ln (x+h) - \ln x = \int_x^{x+h} \frac{dt}{t}$.

I do not have time to finish this.

5. Hello, spammanon!

An interesting challenge . . .

Given: .$\displaystyle f(x) \:=\:x^2\ln x$, find $\displaystyle f'(x)$
. . using the defintion of the derivative: .$\displaystyle f'(x) \;=\;\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

The numerator is: .$\displaystyle f(x+h) - f(x)$

. . . $\displaystyle =\;(x+h)^2\ln(x+h) - x^2\ln(x)$

. . . $\displaystyle = \;(x^2+2xh + h^2)\ln(x+h) - x^2\ln(x)$

. . . $\displaystyle =\;x^2\ln(x+h) + (2xh + h^2)\ln(x+h) - x^2\ln(x)$

. . . $\displaystyle = \;x^2\ln(x+h) - x^2\ln(x) + (2hx+h^2)\ln(x+h)$

. . . $\displaystyle = \;x^2\left[\ln(x+h) - \ln(x)\right] + h(2x+h)\ln(x+h)$

. . . $\displaystyle = \;x^2\ln\left(\frac{x+h}{x}\right) + h(2x+h)\ln(x+h)$

. . . $\displaystyle = \;x^2\ln\left(1 + \frac{h}{x}\right) + h(2x+h)\ln(x+h)$

Divide by $\displaystyle h\!:\;\;\frac{f(x+h)-f(x)}{h}$

. . . $\displaystyle =\;\frac{1}{h}\cdot x^2\ln\left(1 + \frac{h}{x}\right) + (2x+h)\ln(x+h)$

. . . $\displaystyle =\;x \cdot \frac{x}{h}\!\cdot\!\ln\left(1 + \frac{h}{x}\right) + (2x+h)\ln(x+h)]$

. . . $\displaystyle =\;x \cdot\ln\left(1 + \frac{h}{x}\right)^{\frac{x}{h}} + (2x+h)\ln(x+h)$

Take the limit: .$\displaystyle f'(x) \:=\:\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

. . . $\displaystyle =\;\lim_{h\to0} \left[x\!\cdot\!\ln\left(1 + \frac{h}{x}\right)^{\frac{x}{h}} + (2x+h)\ln(x+h)\right]$

. . . $\displaystyle =\;x\cdot\ln\underbrace{\left[\lim_{h\to0}\left(1 + \frac{h}{x}\right)^{\frac{x}{h}}\right]}_{\text{This is }e} \:+ \:\lim_{h\to0}\left[ (2x+h)\ln(x+h)\right]$

. . . $\displaystyle = \;x\ln(e) + 2x\ln(x)$

Therefore: .$\displaystyle f'(x) \;=\;x + 2x\ln(x)$ . . . ta-DAA!

6. ## Thanks to you both

Well done. I had duplicated your work precisely, but the insight about (or recognition of) e was the missing link. Thanks, and I hope you enjoyed it!

7. Originally Posted by spammanon
Well done.
A "Thank You" would be more appropriate

8. He actually said so above with a "Thanks to you both".