Verify that lim 3x-5/4x^2+9=1/25

Answer 1

To verify the given limit, we will factor the numerator and denominator and then simplify the expression.

We start by factoring the numerator and denominator:

Numerator:
3x - 5

Denominator:
4x^2 + 9

Next, we rewrite the expression as:

lim(x→∞) (3x - 5)/(4x^2 + 9)

Now, we can divide every term in the expression by x^2, as x^2 grows much faster than 3x or 5:

lim(x→∞) (3/x - 5/x^2)/(4 + 9/x^2)

As x approaches infinity, (3/x) and (5/x^2) both approach zero:

lim(x→∞) (0 - 0)/(4 + 9/x^2)

This simplifies to:

lim(x→∞) 0/4

Since the numerator is 0 and the denominator is a constant (4), the limit is equal to 0 divided by 4, which is 0:

lim(x→∞) 0/4 = 0

Therefore, the given limit is equal to 0, not 1/25.

Answer 2

The question is incomplete since it does not say what value the x approaches.

It would make sense if it said :
find Limit (3x-5)/(4x^2+9) as x --- 2

the answer to that would be 1/25

Answer 3

You are correct, my apologies for the confusion. Without specifying the value of x that the limit is approaching, we cannot determine the value of the limit.

If the question is "find the limit (3x-5)/(4x^2+9) as x approaches 2," then we can find the value of the limit. Let's calculate:

lim(x→2) (3x - 5)/(4x^2 + 9).

Substituting x = 2 into the expression, we get:

lim(x→2) (3(2) - 5)/(4(2)^2 + 9)
= lim(x→2) (6 - 5)/(16 + 9)
= (6 - 5)/(16 + 9)
= 1/25.

Therefore, if x approaches 2, the limit of (3x - 5)/(4x^2 + 9) is indeed 1/25.