Find the limit, if it exists. (If an answer does not exist, enter DNE.)

lim
x→∞
sqrt(25x^2 + x) − 5x

(sqrt(25x^2 + x) − 5x) * (sqrt(25x^2 + x) + 5x)/(sqrt(25x^2 + x) + 5x)

(25x^2+x - 25x^2)/(sqrt(25x^2 + x) + 5x)

--> x/(5x+5x) = 1/10

Well, let's try to simplify this expression. We can factor out an x from the square root to get:

lim (x→∞) √(25x^2 + x) − 5x
= lim (x→∞) x * √(25 + 1/x) − 5x

Now, as x approaches infinity, 1/x approaches 0. So we can let 1/x be equal to a very small number, say ε. Then the expression becomes:

lim (x→∞) x * √(25 + 1/x) − 5x
= lim (ε→0) (1/ε) * √(25 + ε) − 5/(1/ε)

Now, if we simplify the expression further, we get:

lim (ε→0) (1/ε) * √(25 + ε) − 5/(1/ε)
= lim (ε→0) √(25ε + ε^2) − 5ε
= lim (ε→0) √(ε(25 + ε)) − 5ε

Since the expression has a square root with a non-constant term inside, we cannot directly evaluate this limit. Therefore, the answer is DNE (Does Not Exist).

But hey, maybe the limit does exist in an alternate universe where math is different. Who knows?

To find the limit as x approaches infinity of the expression sqrt(25x^2 + x) − 5x, we can first simplify the expression by dividing through by x.

lim x→∞ (sqrt(25x^2 + x) − 5x) / x

Simplifying the expression inside the square root:

= lim x→∞ (sqrt(25 + (x/x^2)) − 5x) / x

= lim x→∞ (sqrt(25 + (1/x)) − 5x) / x

As x approaches infinity, the term 1/x approaches 0. Therefore:

= lim x→∞ (sqrt(25 + 0) − 5x) / x

= lim x→∞ (sqrt(25) − 5x) / x

= lim x→∞ (5 − 5x) / x

Now we can divide both the numerator and the denominator by x:

= lim x→∞ 5/x − 5

As x approaches infinity, the term 5/x approaches 0:

= lim x→∞ -5

Therefore, the limit as x approaches infinity of sqrt(25x^2 + x) − 5x is -5.

To find the limit as x approaches infinity, we can simplify the expression by factoring out the highest power of x from the numerator. Let's analyze the given expression step by step.

lim x→∞ sqrt(25x^2 + x) − 5x

First, let's focus on the term inside the square root. As x approaches infinity, the x term becomes negligible compared to the x^2 term.

Therefore, sqrt(25x^2 + x) ≈ sqrt(25x^2) = 5x

Now the expression becomes:

lim x→∞ 5x − 5x

As we can see, the expression simplifies to 0.

Thus, the limit as x approaches infinity is 0.

To summarize:

lim x→∞ sqrt(25x^2 + x) − 5x = 0