Question 983102
lim x-> 0  ln(sin|x|)

First you want to plug 0 in to see if the point actually exists.

As you can see, we get ln(0), which is undefined.

Usually for trigonometric limits that involve 0, we think of the squeeze theorem.

Let's set up an inequality here. Start with the trig and build it into our function.

-1 < sin(abs(x)) < 1

ln(-1) < ln(sin(abs(x)) < ln(1)

0 < ln(sin(abs(x)) < 0

so lim x-> 0 ln(sin(abs(x)) = 0


lim x-> - &#8734;  (3+t) / sqrt(1+9t^2)

With this you want to divide everything by the highest degreed term. In this case, that is t.

(3/t + t/t) / (sqrt(1/t^2 + 9t^2/t^2) =   (3/t + 1)/sqrt(1/t^2 + 9)

Plugging in infinity gives us

(0+1)/sqrt(0+9) = 1/sqrt(9) = 1/3