Find the limit. lim x → − π sin x x \lim_{x\rarr-\pi}\frac{\sin x}{x} lim x → − π x s i n x

In this problem, we're asked to find the limit of the sine of x over x as x approaches negative pi. Now upon looking at this problem, this might look really intimidating because not only is it a rational function but it also contains a trig function. But we're going to treat this the way we would any rational function and just first check if our denominator is 0 here or not. Now since I'm looking at x approaching negative pi, if I were to plug that into my denominator then my denominator will just end up being negative pi which is definitely not 0. Now because my denominator is not 0 here that means that I can just proceed in plugging in my x value directly into my function to get my answer here. So this limit is going to be equal to the sine of negative pi plugging in that x value divided by negative pi. Now thinking back to our unit circle the sine of negative pi is just 0 So this ends up being 0 over negative pi which we know is just equal to 0. So the limit of the sine of x over x as x approaches negative pi is 0. Don't get scared away by problems that might look a little intimidating at first. Just take a second and think about it, and you can totally get to an answer. You got this. Thanks for watching. I'll see you in the next one.

Find the limit. lim ⁡ x → − π sin ⁡ x x \lim_{x\rarr-\pi}\frac{\sin x}{x} lim x → − π x s i n x

− 1 π -\frac{1}{\pi} − π 1

22. Limits & Continuity

Finding Limits Algebraically

Precalculus

NEW 22. Limits & Continuity

Finding Limits Algebraically

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Multiple Choice

Find the limit.
$\lim_{x\rarr-\pi}\frac{\sin x}{x}$

$-\frac{1}{\pi}$

$0$

$\pi$

Does not exist

Verified step by step guidance

Step 1: Understand the problem. We need to find the limit of the function $ \frac{\sin x}{x} $ as $ x $ approaches $ -\pi $.

Step 2: Recall the limit properties. The limit of a function $ \frac{f(x)}{g(x)} $ as $ x $ approaches a point can be found if both $ f(x) $ and $ g(x) $ approach a finite value and $ g(x) \neq 0 $.

Step 3: Evaluate the behavior of $ \sin x $ and $ x $ as $ x $ approaches $ -\pi $. Note that $ \sin(-\pi) = 0 $ and $ x $ approaches $ -\pi $.

Step 4: Consider the form $ \frac{0}{-\pi} $. Since $ \sin(-\pi) = 0 $, the numerator approaches 0, and the denominator approaches $ -\pi $, which is a non-zero constant.

Step 5: Conclude that the limit exists and is equal to $ \frac{0}{-\pi} $.