Question

Limit

Answer 1

I love limits

Answer 2

They defined my life for the past year :D…they still do

Answer 3

ln of 0 from the positive side is infinity right?

Answer 4

This is the graph of ln and as you approach 0 from the right you get negative infinity right

Answer 5

so ln(2)/ infinity is 0 because a constant over infinity or negative infinity is 0 right

Answer 6

Then x to the 0 is 1 right

Answer 7

Then you say Wallah and the answer is 1

Answer 8

Am I not suppose to get 1 lol? Well got homework to do so hallah

Answer 9

ik this is a weird one

Answer 10

If you evaluate the limit straight off, we see that \(\large \lim\limits_{x\to 0^+} x^{\ln 2/(1+\ln x)} \rightarrow 0^0\). So we're going to need to rewrite this in a form where we can apply L'Hopital's rule.

Suppose that the limit exists and that \(\large \lim\limits_{x\to0^+} x^{\ln 2/(1+\ln x)} = L\). Taking natural logs of both sides gives us \[\large \begin{aligned} \ln L &= \lim\limits_{x\to 0^+} \ln (x^{\ln 2/(1+\ln x)})\\ &= \lim\limits_{x\to0^+}\dfrac{\ln 2}{1+\ln x}\cdot \ln x\\ &= \ln 2\lim\limits_{x\to0^+}\frac{\ln x}{1+\ln x}\rightarrow \dfrac{\infty}{\infty}.\end{aligned}\]
Now you can apply L'Hopital's rule at this point.

Can you take things from here? :-)

Answer 11

Oh I forgot to plug in 0 for the main x ahahah. Listen to ^ dude

Answer 12

Yes thank you so much! was kind of unsure about taking the ln of the limit but i guess i need to brush up on properties of limits