Question

question 3 1 pts
\\(\lim \limits_{x \to 4^-} \frac{\sqrt{4x} - x}{4 + 3x - x^2}\\)
\\(\bigcirc -\infty\\)
\\(\bigcirc \text{dne}\\)
\\(\bigcirc 1/10\\)
\\(\bigcirc 0\\)

question 4 1 pts
\\(\lim \limits_{x \to \infty} \frac{\sin 5x}{x}\\)
\\(\bigcirc \text{dne}\\)
\\(\bigcirc 0\\)
\\(\bigcirc 5\\)
\\(\bigcirc 1\\)

Explanation:

Question 3

Step 1: Factor the denominator

First, factor the denominator \(4 + 3x - x^2\). We can rewrite it as \(-(x^2 - 3x - 4)\), and then factor the quadratic: \(x^2 - 3x - 4=(x - 4)(x + 1)\), so the denominator is \(-(x - 4)(x + 1)=(4 - x)(x + 1)\).

Step 2: Simplify the numerator and denominator as \(x\to 4^-\)

As \(x\to 4^-\), let's analyze the numerator \(\sqrt{4x}-x\). When \(x = 4\), \(\sqrt{4\times4}-4=4 - 4 = 0\). Now, the denominator \((4 - x)(x + 1)\): as \(x\to 4^-\), \(4 - x\to 0^+\) (since \(x\) is approaching 4 from the left, \(4 - x\) is a small positive number) and \(x + 1\to 5\) (a positive constant).

Now, let's simplify the original function:
\[

$$\begin{align*} \frac{\sqrt{4x}-x}{4 + 3x - x^2}&=\frac{\sqrt{4x}-x}{(4 - x)(x + 1)}\\ &=\frac{\sqrt{4x}-x}{(4 - x)(x + 1)}\times\frac{\sqrt{4x}+x}{\sqrt{4x}+x}\\ &=\frac{4x - x^2}{(4 - x)(x + 1)(\sqrt{4x}+x)}\\ &=\frac{x(4 - x)}{(4 - x)(x + 1)(\sqrt{4x}+x)}\\ &=\frac{x}{(x + 1)(\sqrt{4x}+x)} \quad (x eq 4) \end{align*}$$

\]

Step 3: Evaluate the limit

Now we can substitute \(x = 4\) into the simplified function (since we canceled the non - zero factor \(4 - x\) for \(x
eq 4\)):
\[
\lim_{x\to 4^-}\frac{x}{(x + 1)(\sqrt{4x}+x)}=\frac{4}{(4 + 1)(\sqrt{16}+4)}=\frac{4}{5\times(4 + 4)}=\frac{4}{5\times8}=\frac{1}{10}
\]

Step 1: Recall the boundedness of the sine function

The sine function \(y = \sin t\) is bounded, that is, for any real number \(t\), \(- 1\leqslant\sin t\leqslant1\). In our case, \(t = 5x\), so \(-1\leqslant\sin(5x)\leqslant1\) for all real \(x\).

Step 2: Analyze the behavior of \(\frac{1}{x}\) as \(x\to\infty\)

As \(x\to\infty\), \(\frac{1}{x}\to0\). We can rewrite the function \(\frac{\sin(5x)}{x}\) as \(\sin(5x)\times\frac{1}{x}\).

Step 3: Use the Squeeze Theorem

We know that \(-1\leqslant\sin(5x)\leqslant1\). Multiply each part of the inequality by \(\frac{1}{x}\) (for \(x>0\), the inequality direction remains the same):
\[
-\frac{1}{x}\leqslant\frac{\sin(5x)}{x}\leqslant\frac{1}{x}
\]
Now, we find the limits of the left - hand side and the right - hand side as \(x\to\infty\):
\(\lim_{x\to\infty}-\frac{1}{x}=0\) and \(\lim_{x\to\infty}\frac{1}{x}=0\).
By the Squeeze Theorem, \(\lim_{x\to\infty}\frac{\sin(5x)}{x}=0\).

Answer:

\(\frac{1}{10}\)

Question 4