How do I solve the following limit?

Question

lim sin(x)^x

x→0+

For some context I'm in AP Calculus AB and I just covered L'Hopital's Rule, so I'm allowed to use it where applicable. I've tried using trig identities and logarithms to change the expression, but this has not gotten me any closer to the answer. What do I do?

Answer 1

lim_x→0+ (sinx)^x has the indeterminate form 0⁰.

Let y = (sinx)^x. Then lny = ln(sinx)^x = x ln(sinx)

lim_x→0+ lny = lim_x→0+ (x ln(sinx)) = lim_x→0+[ln(sinx) / (1/x)] which has the form -∞/∞

Applying L'Hopital's Rule, the limit on the line above is equivalent to:

lim_x→0+ [ ((cosx) / (sinx)) / (-1/x²)] = lim_x→0+[ (-x/sinx)(x)(cosx)] = (-1)(0)(1) = 0

So, lim_x→0+(lny) = 0

Therefore, lim_x→0+(sinx)^x = lim_x→0+(y) = lim_x→0+(e^lny) = e⁰ = 1

Answer 2

Hi Ben,

Use L'Hopital's rule here

lim_x→₀ (sinx/x)=lim_x→₀[(sinx)^'/(x^')]=

=lim_x→₀[cos(x)/1]=cos(0)=1

B/c If lim_x→₀ exists, then

lim_x→₀₊=lim_x→_0-=lim_x→₀=1

Therefore, lim_x→₀₊(sinx/x)=1

Hope it helps

Best

Dr.Ariel B.