# How do you evaluate the limit #(abs(x+2)-2)/absx# as x approaches #0#?

The limit:

does not exist since the right and left limit are different.

Evaluate separately:

We can conclude that:

does not exist since the right and left limit are different.

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

To evaluate the limit (abs(x+2)-2)/absx as x approaches 0, we can consider the cases when x is positive and when x is negative.

When x is positive, the absolute value of x+2 is equal to x+2, and the absolute value of x is equal to x. Therefore, the expression becomes (x+2-2)/x, which simplifies to x/x, resulting in 1.

When x is negative, the absolute value of x+2 is equal to -(x+2), and the absolute value of x is equal to -x. Thus, the expression becomes (-(x+2)-2)/(-x), which simplifies to (-x-2-2)/(-x), resulting in (-x-4)/(-x), which further simplifies to (x+4)/x.

Since the limit is evaluated as x approaches 0, we can consider the limit from both the positive and negative sides. In both cases, the limit evaluates to 1.

Therefore, the limit of (abs(x+2)-2)/absx as x approaches 0 is equal to 1.

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

- How do you evaluate the limit #cos((pix)/3)# as x approaches #2#?
- How do you find the limit of #(arctan(5x))/(sin(7x))# as x approaches 0?
- How do you find the limit #lnx/sqrtx# as #x->oo#?
- What is the limit of #(x-1)(x-2)(x-3)(x-4)(x-5)/(5x-1)^5# as x goes to infinity?
- How do you evaluate the limit #(x^2-x-6)/(x+2)# as x approaches #oo#?

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