# How do you find the limit of #cosx/(x^2+9)# as #x->0#?

The function is continuous in

By signing up, you agree to our Terms of Service and Privacy Policy

To find the limit of cosx/(x^2+9) as x approaches 0, we can substitute 0 into the expression. This gives us cos(0)/(0^2+9), which simplifies to 1/9. Therefore, the limit of cosx/(x^2+9) as x approaches 0 is 1/9.

By signing up, you agree to our Terms of Service and Privacy Policy

To find the limit of (\frac{\cos(x)}{x^2 + 9}) as (x) approaches (0), you can directly substitute (x = 0) into the expression.

Substituting (x = 0), we get:

[ \lim_{x \to 0} \frac{\cos(x)}{x^2 + 9} = \frac{\cos(0)}{0^2 + 9} = \frac{1}{9} ]

So, the limit of (\frac{\cos(x)}{x^2 + 9}) as (x) approaches (0) is (\frac{1}{9}).

By signing up, you agree to our Terms of Service and Privacy Policy

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

- What is the limit of # ((1 + 3 x) / (2 + 4 x^2 + 2 x^4))^3 # as x approaches 1?
- How to resolve this limit from Vector Calculus?
- How do you find the limit of # (5^t-3^t)/t# as t approaches 0?
- How do you find the limit #sqrt(x^2+x)-sqrt(x^2-x)# as #x->oo#?
- How do you evaluate the limit #sqrt(x-2)# as x approaches #5#?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7