Question

use L'Hôpital's rule to solve lim(x->0+) x^2x

Answer 1

\[\lim_{x \rightarrow 0⁺}x²^x\]

that's x to the power 2x

Answer 2

so it's quite clearly of the form 0^0 and therefore it's ok to use L'Hôpital's rule

Answer 3

but the 1st iteration gives 0^-1

Answer 4

\[\large\lim_{x \rightarrow 0^+}x^{2x} \]

Answer 5

\[\large\lim_{x \rightarrow 0^+}x^{2x}=\large\lim_{x \rightarrow 0^+}e^{\ln x^{2x}}=\large\lim_{x \rightarrow 0^+}e^{2x\ln x}=\]

Answer 6

@UnkleRhaukus Darn.

Answer 7

so then the derivative is lim(x->0) ln(x)*e^2x*ln(x)=ln(x)*x^2x

Answer 8

\[\large\lim_{x \rightarrow 0^+}e^{2x\ln x}=e^{~\lim\limits_{x \rightarrow 0^+}2x\ln x}=\]

Answer 9

:')

Answer 10

\[\large e^{~\lim\limits_{x \rightarrow 0^+}2x\ln x}=e^{~2\lim\limits_{x \rightarrow 0^+}\frac{\ln x}{1/x}}=e^{~2\lim\limits_{x \rightarrow 0^+}\frac{1/x}{1/-x^2}}=e^{~2\lim\limits_{x \rightarrow 0^+}-x}\]

Answer 11

so there is no need to take a derivative and the limit as x approaches 0 is 1?

Answer 12

i've used L'Hôp twice

Answer 13

ok I see where you used L'Hôpital's. I'm still a little confused concerning some of the rules (1. why is it ok to put the limit to the exponent? 2. why is it ok to differentiate only the exponent? 3. What is the rule used in the 1st differentiation?). Thanks for the help anyways @UnkleRhaukus I'll try to wrap my head around it

Answer 14

1. the exponential function is continuos

Answer 15

2.

\[{\lim\limits_{x \rightarrow 0^+}x\ln x}\]\[\qquad\qquad\qquad \text{let } {x=1/X}\]\[-{\lim\limits_{X \rightarrow \infty}\frac{\ln\tfrac 1X}{X}}\]
\[-{\lim\limits_{X \rightarrow \infty}\frac{\tfrac 1X}{1}}\\-\lim\limits_{X \rightarrow \infty}\frac1X\\-\lim\limits_{x \rightarrow 0^+}x\]

Answer 16

make sense now?

Answer 17

:D not exactly. I got point 1.

Answer 18

3. Why does\[2x \ln(x)\] become\[2*\frac{ \ln(x) }{ 1/x }\] and not\[2*\ln(x)+\frac{ 1 }{ x }*2x=2*\ln(x)+2\] ?

Answer 19

well this is L'Hôp, we had the form \[m\times n \to (0)\times (-\infty)\]so we differentiate \(m\) and \(n\) separately

(we are not using the product rule here)