Find the limit as x approaches 0 of (sin x)/(x + tan x)
To find the limit as x approaches 0 of (sin x)/(x + tan x), we can use L'Hôpital's rule.
L'Hôpital's rule states that if we have a limit of the form f(x)/g(x) as x approaches a specific value and both f(x) and g(x) approach 0 or ±∞, then we can differentiate the numerator and denominator repeatedly until we can evaluate the limit.
Let's apply L'Hôpital's rule to the given problem:
Step 1: Differentiate the numerator and denominator:
The derivative of sin(x) is cos(x).
The derivative of x + tan(x) is 1 + sec^2(x).
Step 2: Substitute these derivatives back into the limit expression:
lim (x->0) [cos(x)] / [1 + sec^2(x)]
Step 3: Simplify the expression:
As x approaches 0, cos(x) approaches 1, and sec^2(x) also approaches 1 (since sec^2(x) is equal to 1 + tan^2(x)).
lim (x->0) [1] / [1 + 1]
= 1/2
Therefore, the limit as x approaches 0 of (sin x)/(x + tan x) is 1/2.