# [SOLVED] Difficult limit

• Feb 15th 2008, 11:13 AM
spammanon
[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.
• Feb 15th 2008, 11:54 AM
ThePerfectHacker
Quote:

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.
• Feb 15th 2008, 11:58 AM
spammanon
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.
• Feb 15th 2008, 12:58 PM
ThePerfectHacker
$\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.
• Feb 15th 2008, 01:13 PM
Soroban
Hello, spammanon!

An interesting challenge . . .

Quote:

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!

• Feb 15th 2008, 03:30 PM
spammanon
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!
• Feb 15th 2008, 03:35 PM
bobak
Quote:

Originally Posted by spammanon
Well done.

A "Thank You" would be more appropriate ;)
• Feb 15th 2008, 03:37 PM
Krizalid
He actually said so above with a "Thanks to you both".

:D