# How do you evaluate the limit #((x+5)^2-25)/x# as x approaches #0#?

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

To evaluate the limit ((x+5)^2-25)/x as x approaches 0, we can substitute 0 into the expression. This gives us ((0+5)^2-25)/0, which simplifies to (5^2-25)/0. Further simplifying, we have (25-25)/0, which is 0/0. This is an indeterminate form. To evaluate further, we can factor the numerator as (5-5)(5+5)/0, which becomes 0(10)/0. Simplifying, we get 0/0 again. Since we have an indeterminate form, we need to use additional techniques such as factoring, rationalizing, or L'Hôpital's rule to evaluate the limit.

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.

- How do you find the limit #sin(2x)/ln(x+1)# as #x->0#?
- How do you evaluate the limit #(root3x-1)/(sqrtx-1)# as x approaches #1#?
- What are the discontinuities of #f(x) = (x^4 - 1)/(x-1)#?
- How do you find the limit of #e^sqrtx/(sqrt(e^x+1)# as #x->oo#?
- How do you find #lim sqrt(x+2)-sqrtx# as #x->oo#?

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