answersLogoWhite

0

The limit as x approaches zero of sin(x) over x can be determined using the squeeze theorem.

For 0 < x < pi/2, sin(x) < x < tan(x)

Divide by sin(x), and you get 1 < x/sin(x) < tan(x)/sin(x)

That is the same as 1 < x/sin(x) < 1/cos(x)

But the limit as x approaches zero of 1/cos(x) is 1,

so 1 < x/sin(x) < 1

which means that the limit as x approaches zero of x over sin(x) is 1, and that also means the inverse; the limit as x approaches zero of sin(x) over x is 1.

You can also solve this using deriviatives...

The deriviative d/dxx is 1, at all points. The deriviative d/dxsin(x) at x=0 is also 1.

This means you have the division of two functions, sin(x) and x, at a point where their slope is the same, so the limit reduces to 1 over 1, which is 1.

User Avatar

Wiki User

15y ago

Still curious? Ask our experts.

Chat with our AI personalities

RafaRafa
There's no fun in playing it safe. Why not try something a little unhinged?
Chat with Rafa
CoachCoach
Success isn't just about winning—it's about vision, patience, and playing the long game.
Chat with Coach
JudyJudy
Simplicity is my specialty.
Chat with Judy

Add your answer:

Earn +20 pts
Q: Evaluate limit x tends to 0 sinx x?
Write your answer...
Submit
Still have questions?
magnify glass
imp