# How do you use the epsilon delta definition to find the limit of #x^2 cos(1/x)# as x approaches #0#?

The epsilon-delta formulation of the definition of limit states that:

if:

which proves that:

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

To use the epsilon-delta definition to find the limit of (x^2 \cos(1/x)) as (x) approaches 0, we need to show that for any given epsilon ((\varepsilon)) greater than 0, there exists a delta ((\delta)) greater than 0 such that if (0 < |x - 0| < \delta), then (|x^2 \cos(1/x) - L| < \varepsilon), where (L) is the limit we are trying to find.

First, let's simplify the expression (x^2 \cos(1/x)). Since the cosine function is bounded between -1 and 1, we can say that (|x^2 \cos(1/x)| \leq |x^2|) for all (x). Therefore, we can focus on finding the limit of (x^2) as (x) approaches 0.

Now, let's consider the inequality (|x^2 - L| < \varepsilon). We want to find a suitable (\delta) such that if (0 < |x - 0| < \delta), then (|x^2 - L| < \varepsilon).

To do this, we can manipulate the inequality as follows:

(|x^2 - L| < \varepsilon) (-\varepsilon < x^2 - L < \varepsilon) (L - \varepsilon < x^2 < L + \varepsilon)

Since we are interested in the limit as (x) approaches 0, we can assume that (|x|) is small. Therefore, we can choose (\delta) such that (\delta < 1) (for example).

Now, if we choose (\delta = \sqrt{L + \varepsilon} - \sqrt{L - \varepsilon}), we can show that if (0 < |x - 0| < \delta), then (|x^2 - L| < \varepsilon).

By substituting (\delta) into the inequality, we get:

(L - \varepsilon < x^2 < L + \varepsilon) (L - \varepsilon < x^2 < L + \varepsilon) (L - \varepsilon < (\sqrt{L + \varepsilon} - \sqrt{L - \varepsilon})^2 < L + \varepsilon)

Expanding the square and simplifying, we have:

(L - \varepsilon < L + \varepsilon - 2\sqrt{(L + \varepsilon)(L - \varepsilon)} < L + \varepsilon)

Since (\delta < 1), we can conclude that if (0 < |x - 0| < \delta), then (|x^2 - L| < \varepsilon).

Therefore, the limit of (x^2 \cos(1/x)) as (x) approaches 0 is (L).

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 of #(2+x)^(1/x)# as x approaches #0^-#?
- How do you prove that the limit of #(x^2 - 4x + 5) = 1# as x approaches 2 using the epsilon delta proof?
- How do I find #lim_((x,y) to (5,4)) e^sqrt (3x^2+2y^2)#, if it exists?
- How do you find the limit of #n/(3n-1)# as n approaches infinity?
- How do you use a graphing calculator to find the limit of #x^2-5x# as x approaches -1?

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