# How do you prove that the limit #(x^2+x-4)=8# as x approaches 3 using the formal definition of a limit?

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

To prove that the limit of the function (x^2+x-4) is 8 as x approaches 3 using the formal definition of a limit, we need to show that for any positive value ε, there exists a positive value δ such that if 0 < |x - 3| < δ, then |(x^2+x-4) - 8| < ε.

Let's proceed with the proof:

Given the function f(x) = x^2 + x - 4, we want to show that for any ε > 0, there exists a δ > 0 such that if 0 < |x - 3| < δ, then |(x^2+x-4) - 8| < ε.

First, let's simplify the expression |(x^2+x-4) - 8|:

|(x^2+x-4) - 8| = |x^2 + x - 12|

Now, we can factor the expression x^2 + x - 12:

x^2 + x - 12 = (x - 3)(x + 4)

Next, we can rewrite the absolute value expression using the factored form:

|(x^2+x-4) - 8| = |(x - 3)(x + 4)|

Since we are interested in the behavior of the function as x approaches 3, we can assume that 0 < |x - 3| < 1 (we will use this assumption later).

Now, let's consider the expression |(x - 3)(x + 4)|:

|(x - 3)(x + 4)| = |x - 3||x + 4|

Since we have assumed that 0 < |x - 3| < 1, we can also assume that 0 < x - 3 < 1, which implies -1 < x - 3 < 0.

Using this assumption, we can establish a bound for |x + 4|:

-1 < x - 3 < 0 3 < x < 4 7 < x + 4 < 8

Therefore, we can conclude that |x + 4| < 8.

Now, let's choose δ = min{1, ε/8}.

If 0 < |x - 3| < δ, then we have:

0 < |x - 3| < 1 |x - 3||x + 4| < δ|x + 4| < (ε/8) * 8 = ε

Hence, we have shown that for any ε > 0, there exists a δ > 0 such that if 0 < |x - 3| < δ, then |(x^2+x-4) - 8| < ε. Therefore, the limit of the function (x^2+x-4) as x approaches 3 is indeed 8, as desired.

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.

- Evaluate #lim _(x-> oo) (sinxcosx)/(3x) # ?
- How do you find the Limit of # (2x+3)[ln(x^2-x+1)-ln(x^2+x+1)]# as x approaches infinity?
- How do you find the limit of #(x^n-1) / (x-1)# as x approaches 1?
- Evaluate the limit # lim_(h rarr 0) (1-cos h)/h^2 #?
- For what values of x, if any, does #f(x) = 1/((x^2-4)(x+3)(x-2)) # have vertical asymptotes?

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