Limit solving: Find lim x->inf ((sqrt(x))^lnx) / e^x

bobb17

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Hello, I want to solve the limit

lim x->inf ((sqrt(x))^lnx) / e^x

My thought was to apply the natural logarithm function on both the numerator and the denominator to simplify the exponent part on the numerator then apply L'hopital rule:

= lim x->inf ln[((sqrt(x))^lnx)] / ln[e^x] = lim x->inf (lnx)(ln(sqrtx)) / x = ... = 0.

Can you tell me if applying the natural logarithm function can be done this way?
Thanks
 
I want to solve the following limit:

. . . . .\(\displaystyle \displaystyle \lim_{x \rightarrow \infty}\, \dfrac{\left(\sqrt{ x\,}\right)^{\ln(x)}}{e^x}\)

My thought was to apply the natural logarithm function on both the numerator and the denominator to simplify the exponent part on the numerator then apply L'Hopital's rule:

. . . . .\(\displaystyle \displaystyle =\, \lim_{x\rightarrow \infty}\, \dfrac{\ln\big(\left(\sqrt{x\,}\right)^{\ln(x)} \big)}{\ln(e^x)}\)

. . . . .\(\displaystyle \displaystyle =\, \lim_{x \rightarrow \infty}\, \dfrac{\ln(x)\, \cdot\, \ln\left(\sqrt{x\,}\right)}{x}\)

. . . . .\(\displaystyle \displaystyle =\, ...\, =\, 0\)

Can you tell me if applying the natural logarithm function can be done this way?
Isn't the following the actual process for doing log limits?

. . . . .\(\displaystyle \displaystyle \mbox{Let }\, y\, =\, \lim_{x \rightarrow \infty}\, f(x).\, \mbox{ Then }\, \ln(y)\, =\, \ln\bigg(\lim_{x \rightarrow \infty}\, f(x)\bigg)\, =\, \lim_{x \rightarrow \infty}\big(\ln(f(x))\big)\)

And, since:

. . . . .\(\displaystyle \ln\left(\dfrac{a}{b}\right)\, \neq\, \dfrac{\ln(a)}{\ln(b)}\)

...I don't think one can make the conclusion that you have. Instead, I think the rule is as follows:

. . . . .\(\displaystyle \ln\left(\dfrac{a}{b}\right)\, =\, \ln(a)\, -\, \ln(b)\)

Where does this lead? ;)
 
Ok so how would someone solve this limit?
One approach is to use basic properties of exponents and logarithms (like the one stapel suggested you try, as one example), to re-express the given function in a form that's easier to work with. Experience dealing with indeterminant forms is also helpful.

For example, were you to have the following limit, could you solve it?

limit as x→∞ of e^[1/2*ln(x)^2 - x]
 
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