Question

1. (4 points; 2 points each) for each question below, identify the form and use lhôpitals rule if it applies. choose 2. last one is worth 4 points.

a. (2 points) \\( \lim _ { x \
ightarrow 1 } \frac { x ^ { 2 } + 3 x - 5 } { x - 1 } \\)
b. (2 points) \\( \lim _ { x \
ightarrow 0 } \frac { \sin ( 2 x ) } { 3 x } \\)
c. (2 points) \\( \lim _ { x \
ightarrow 0 } \frac { e ^ { x } - 1 } { \sin ( x ) } \\)
d. (4 points) \\( \lim _ { x \
ightarrow 0 } ( \frac { 1 } { x } - \frac { 1 } { \sin ( x ) } ) \\)
(hint: rewrite the form and apply lhôpitals rule twice)

Explanation:

Step1: Identify the form for part b

For \(\lim_{x
ightarrow0}\frac{\sin(2x)}{3x}\), substituting \(x = 0\) gives \(\frac{\sin(0)}{0}=\frac{0}{0}\) form.

Step2: Apply L'Hôpital's rule

Differentiate numerator and denominator. The derivative of \(\sin(2x)\) is \(2\cos(2x)\) and derivative of \(3x\) is \(3\). So, \(\lim_{x
ightarrow0}\frac{\sin(2x)}{3x}=\lim_{x
ightarrow0}\frac{2\cos(2x)}{3}\)

Step3: Evaluate the limit

Substitute \(x = 0\) into \(\frac{2\cos(2x)}{3}\). We get \(\frac{2\cos(0)}{3}\), since \(\cos(0)=1\), then \(\frac{2\times1}{3}=\frac{2}{3}\)

Step1: Identify the form for part c

For \(\lim_{x
ightarrow0}\frac{e^{x}-1}{\sin(x)}\), substituting \(x = 0\) gives \(\frac{e^{0}-1}{\sin(0)}=\frac{1 - 1}{0}=\frac{0}{0}\) form.

Step2: Apply L'Hôpital's rule

Differentiate numerator and denominator. The derivative of \(e^{x}-1\) is \(e^{x}\) and derivative of \(\sin(x)\) is \(\cos(x)\). So, \(\lim_{x
ightarrow0}\frac{e^{x}-1}{\sin(x)}=\lim_{x
ightarrow0}\frac{e^{x}}{\cos(x)}\)

Step3: Evaluate the limit

Substitute \(x = 0\) into \(\frac{e^{x}}{\cos(x)}\). We get \(\frac{e^{0}}{\cos(0)}\), since \(e^{0}=1\) and \(\cos(0)=1\), then \(\frac{1}{1}=1\)

Answer:

For part b: \(\frac{2}{3}\)
For part c: \(1\)