6. $ \lim _ { x \
ightarrow 0 } \frac { x ^ { 4 } - 5 x ^ { 3 } } { x ^ { 3 } + 5 x ^ { 7 } } $

Question

Accepted Answer

# Explanation:
## Step1: Factor out the highest - power of $x$ in the numerator and denominator
The highest - power of $x$ in the numerator $x^{4}-5x^{3}$ is $x^{3}$, and in the denominator $x^{3}+5x^{7}$ is $x^{3}$.
$$
\begin{align*}
\lim_{x
ightarrow0}\frac{x^{4}-5x^{3}}{x^{3}+5x^{7}}&=\lim_{x
ightarrow0}\frac{x^{3}(x - 5)}{x^{3}(1 + 5x^{4})}\
\end{align*}
$$
## Step2: Cancel out the common factor $x^{3}$
Since $x
eq0$ when taking the limit as $x
ightarrow0$ (we are approaching $0$ but not equal to $0$), we can cancel $x^{3}$ from the numerator and denominator.
$$
\begin{align*}
\lim_{x
ightarrow0}\frac{x^{3}(x - 5)}{x^{3}(1 + 5x^{4})}&=\lim_{x
ightarrow0}\frac{x - 5}{1+5x^{4}}
\end{align*}
$$
## Step3: Substitute $x = 0$ into the simplified function
We use the direct - substitution property of limits $\lim_{x
ightarrow a}f(x)=f(a)$ for a continuous function $y = f(x)$ at $x=a$. The function $y=\frac{x - 5}{1+5x^{4}}$ is a rational function and is continuous at $x = 0$.
$$
\begin{align*}
\lim_{x
ightarrow0}\frac{x - 5}{1+5x^{4}}&=\frac{0-5}{1+5\times0^{4}}\
&=\frac{- 5}{1}\
&=-5
\end{align*}
$$

# Answer:
$-5$

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QUESTION IMAGE

Question

Explanation:

Step1: Factor out the highest - power of \(x\) in the numerator and denominator

Step2: Cancel out the common factor \(x^{3}\)

Step3: Substitute \(x = 0\) into the simplified function

Answer: