Question

question 1 1 pts
\lim_{x \to 4^+} \frac{2x - 5}{4 - x}
options: dne, -\infty, 0, \infty
question 2 1 pts
\lim_{x \to 1} \frac{x - 1}{\sin(x^2 - 1)}
options: dne, 0, 2, 1/2

Explanation:

Question 1

Step1: Analyze numerator and denominator as \( x \to 4^+ \)

For the numerator \( 2x - 5 \), when \( x \to 4^+ \), substitute \( x = 4 \): \( 2(4)-5 = 8 - 5 = 3 \) (a positive constant).
For the denominator \( 4 - x \), when \( x \to 4^+ \), \( x > 4 \), so \( 4 - x < 0 \) and \( 4 - x \to 0^- \) (approaches 0 from the negative side).

Step2: Determine the limit behavior

A positive numerator divided by a denominator approaching \( 0^- \) (negative and near 0) means the fraction approaches \( -\infty \) (since \( \frac{\text{positive}}{\text{negative small}} \to -\infty \)).

Step1: Use the limit identity \( \lim_{u \to 0} \frac{\sin u}{u} = 1 \)

First, factor the denominator: \( x^2 - 1 = (x - 1)(x + 1) \). Rewrite the limit:
\( \lim_{x \to 1} \frac{x - 1}{\sin((x - 1)(x + 1))} \)

Let \( u = (x - 1)(x + 1) \). As \( x \to 1 \), \( u \to 0 \). We can rewrite the numerator and denominator:
\( \frac{x - 1}{\sin(u)} = \frac{x - 1}{\sin((x - 1)(x + 1))} = \frac{1}{x + 1} \cdot \frac{(x - 1)(x + 1)}{\sin((x - 1)(x + 1))} \)

Step2: Apply the limit identity

Now, split the limit:
\( \lim_{x \to 1} \frac{1}{x + 1} \cdot \lim_{x \to 1} \frac{(x - 1)(x + 1)}{\sin((x - 1)(x + 1))} \)

For the first limit: \( \lim_{x \to 1} \frac{1}{x + 1} = \frac{1}{1 + 1} = \frac{1}{2} \).
For the second limit, let \( v = (x - 1)(x + 1) \), so \( \lim_{v \to 0} \frac{v}{\sin v} = 1 \) (by the identity \( \lim_{u \to 0} \frac{\sin u}{u} = 1 \), so \( \lim_{u \to 0} \frac{u}{\sin u} = 1 \)).

Multiply the two limits: \( \frac{1}{2} \cdot 1 = \frac{1}{2} \).

Answer:

\( -\infty \) (corresponding to the option \( -\infty \))

Question 2