Using L'Hopital's rule to evaluate limit

Question

How would I evaluate limx→0 (x + cos(x))^(1/x) using l'hopital's rule? I am using the rule of logs to simplify the expression to (1/x)*ln(x+cos(x)), but I am unsure of how to go on from there.

Answer 1

You've taken a good first step

L = lim_x->0 ( x + cos x )^1/x

ln L = lim_x->0 [ ln ( x + cos x ) ] / x

Now, if you try to evaluate the limit by directly plugging in x=0, on the RHS you get the Indeterminate Form 0/0.

[ ln ( 0 + 1) ] / 0 = 0/0

This tells you to use L'Hopital's Rule.

L'Hopital's Rule says that if we get a form like 0/0, we can take the derivative of the numerator and denominator, and if we get a meaningful number, then we have the limit.

(Not to be confused with the quotient rule for derivatives!)

Numerator:

f(x) = ln( x + cos x )

f'(x) = (1 - sin x)/(x + cos x)

(Chain rule!)

Denominator:

g(x) = x

g'(x) = 1

Now use L'Hopital's Rule

lim_x->0 f(x)/g(x) = lim_x->0 f'(x)/g'(x)

lim_x->0 ln ( x + cos x ) / x = lim_x->0 (1 - sin x)/(x + cos x)

Now, if we evaluate the above expression at the limit x=0, we get a meaningful number (0/0 is ambiguous, since things can approach 0 at different speeds!)

x->0:

(1 - sin 0)/(0 + cos 0) = 1/1

Since 1/1 makes sense, we have taken a limit successfully.

Remember, we used the natural log to change things around, so we'll have to work our way back from that via exponentiation

L = lim_x->0 ( x + cos x )^1/x

ln L = lim_x->0 [ ln ( x + cos x ) ] / x = 1

ln L = 1

L = e¹=e