limit x^2*(3^(1/x)-3^(1/x+1) for x to infinity

[studentnr001]

I'm creating bald spots from scratching my head, figuring this puzzle out.

I have to calculate the limit for x²*(3^1/x-3^1/(x+1) with x going to infinity.
Wolfram Alpha correctly calculates its answer: ln (3), ,but doesn't offer step-by-step solutions (symbolab doesn't even compute)

I know I have to apply the rule \(\displaystyle \displaystyle \, \lim_{x \rightarrow a}\, f(x)\, =\, \lim_{x \rightarrow a}\, e^{\ln(f(x))}\,\) but that gives me e^{2lnx-ln(3^(1/x)-3^(1/(x+1)} which leads to infinity. I can't seem to figure how to reorder it to achieve ln(3).

I do have figured out to eliminate the x² by applying \(\displaystyle \displaystyle \, \lim_{x \rightarrow a}\, f(x)\, =\, \lim_{x \rightarrow a}\, \ln\left(e^{f(x)}\right)\,\) so ln e^{x^2*(3^(1/x)-3^(1/(x+1)))} = ln e^{x^2*3^1/x}/e^{x^2*3^(1/(x+1))} = ln e^{3^(1/x-1/(x+1)))} = 3^1/x-3^1/(x+1). The limit of this equation gives me 0 for x to infinity.

Please guide me the right way to solve this puzzle.

 

[Steven G]

I'm creating bald spots from scratching my head, figuring this puzzle out.

I have to calculate the limit for x²*(3^1/x-3^1/(x+1) with x going to infinity.
Wolfram Alpha correctly calculates its answer: ln (3), ,but doesn't offer step-by-step solutions (symbolab doesn't even compute)

I know I have to apply the rule \(\displaystyle \displaystyle \, \lim_{x \rightarrow a}\, f(x)\, =\, \lim_{x \rightarrow a}\, e^{\ln(f(x))}\,\) but that gives me e^{2lnx-ln(3^(1/x)-3^(1/(x+1)} which leads to infinity. I can't seem to figure how to reorder it to achieve ln(3).

I do have figured out to eliminate the x² by applying \(\displaystyle \displaystyle \, \lim_{x \rightarrow a}\, f(x)\, =\, \lim_{x \rightarrow a}\, \ln\left(e^{f(x)}\right)\,\) so ln e^{x^2*(3^(1/x)-3^(1/(x+1)))} = ln e^{x^2*3^1/x}/e^{x^2*3^(1/(x+1))} = ln e^{3^(1/x-1/(x+1)))} = 3^1/x-3^1/(x+1). The limit of this equation gives me 0 for x to infinity.

Please guide me the right way to solve this puzzle.

No need to use logs! Try L'Hopital's rules!

 

[studentnr001]

No need to use logs! Try L'Hopital's rules!

Forgive me for my densenesss, the eurekamoment hasn't hit me yet.

In what way can I rearrange the infinity*0 result to a fruitful 0/0 or infinity/infinity?

I can separate the x² by x²*(3^1/x-3^1/(x+1))/(x²*1/x²) and after applying Hôpital I get: (-1/x²*3^(1/x)+3^1/(x+1)*1/(x+1)²)*-x³/2=x/2*3^1/x-x³/(x+1)²*3^1/(x+1)

Which is still a dead end (infinity-infinity).

 

[Steven G]

Forgive me for my densenesss, the eurekamoment hasn't hit me yet.

In what way can I rearrange the infinity*0 result to a fruitful 0/0 or infinity/infinity?

I can separate the x² by x²*(3^1/x-3^1/(x+1))/(x²*1/x²) and after applying Hôpital I get: (-1/x²*3^(1/x)+3^1/(x+1)*1/(x+1)²)*-x³/2=x/2*3^1/x-x³/(x+1)²*3^1/(x+1)

Which is still a dead end (infinity-infinity).

Change multiplication to division.