Question

Lim x³+12²-5x/5x
X=>0

Answer 1

To evaluate the limit as \( x \) approaches 0 for the expression \(\frac{x^3 + 12^2 - 5x}{5x}\), we start by substituting \( x = 0 \) directly into the expression:

\[
\frac{0^3 + 12^2 - 5(0)}{5(0)} = \frac{0 + 144 - 0}{0} = \frac{144}{0}
\]

This results in an undefined expression since division by zero is not allowed. Therefore, we need to analyze the limit further.

Let's first simplify the expression:

\[
\frac{x^3 + 144 - 5x}{5x}
\]

We can break this fraction into separate terms:

\[
= \frac{x^3}{5x} + \frac{144}{5x} - \frac{5x}{5x}
\]
\[
= \frac{x^2}{5} + \frac{144}{5x} - 1
\]

Now we will analyze the limit of each term as \( x \) approaches 0:

1. \(\frac{x^2}{5} \to 0\) as \( x \to 0\)
2. \(\frac{144}{5x} \to \infty\) as \( x \to 0\) (since it approaches \( +\infty \))
3. \(-1\) remains \(-1\)

The term \(\frac{144}{5x}\) dominates the other terms as \( x \) approaches 0. Thus, we conclude that:

\[
\lim_{x \to 0} \left( \frac{x^2}{5} + \frac{144}{5x} - 1 \right) = \infty
\]

Therefore, the limit does not exist in the finite sense and tends towards \( +\infty \):

\[
\lim_{x \to 0} \frac{x^3 + 12^2 - 5x}{5x} = +\infty
\]