This post comes mostly from the youtube video by BlackPenRedPen found here: https://www.youtube.com/watch?v=89d5f8WUf1Y&t=3s
This in turn comes from Brilliant.com – details and links can be found in the original video
In this post we will have a look at a complicated-looking limit that has an interesting solution. Here it is:
This looks pretty daunting – but we will break the solution down into sections:
- taking the logarithms and rearranging
- recognising something familiar
- finding the numerical value
Step 1: Taking the Logarithm
The first step here is to take the logarithm, a generally useful trick when applying limits. First we assign the variable L to the limit (so that we can solve for it in the end). Now lets do some algebra:
Noting that the natural logarithm is a continuous function and therefore we can take the limit outside of the function:
Next we can use the logarithm laws to bring down the exponent:
Alright, now we have taken the logarithm, step 1 is complete.
Step 2: Rearranging the expression and recognising the form of the integral
Now we turn our attention to the expression . Breaking this down a bit more we can get closer to solving this. We know that and that .
So we can rewrite:
Then using log laws and changing the order we have:
Next we put this back into our full expression to get:
But this looks a lot like the definition of the Riemann integral:
where we have split the interval into n increments, and .
So we can therefore rewrite our expression above as:
Now all that we need to do is find the value of the integral
Step 3: Evaluating the integral and solving for L
The last thing we need to do is solve this integral – we will do this by parts.
The 2 functions and their derivatives () are as follows:
which gives us:
Looking at the first term we have:
Now to handle this limit, which is in the indeterminate form – as usual we will rewrite it and then use L’Hospital’s rule to evaluate it:
Using L’Hospital’s rule and differentiating the numerator and the denominator and taking their respective limits we get:
This gives us the first term as 0.
Clearly the second term, ,
Therefore our integral evaluates to -1.
Now all that remains is to rearrange our equation and solve for
and that’s it, we are done:
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