Question

二、find the limits:
(1) \\( \lim _ { x \
ightarrow 0 } \frac { \sin 3 x } { \sin x } \\)
(2) \\( \lim _ { x \
ightarrow 0 } \frac { x } { x + 1 } \\)
(3) \\( \lim _ { x \
ightarrow 0 ^ { + } } \frac { \sin 4 x ^ { 2 } } { x } \\)
(4) \\( \lim _ { x \
ightarrow \infty } \frac { x ^ { 2 } - 1 } { \ln x } \\)
(5) \\( \lim _ { x \
ightarrow \infty } \frac { x ^ { 2 } } { e ^ { x } } \\)
(6) \\( \lim _ { x \
ightarrow \infty } \frac { x ^ { 2 } + 2 } { 7 x ^ { 2 } } \\)

Explanation:

Step1: Apply the limit formula

For \(\lim_{x
ightarrow0}\frac{\sin ax}{\sin bx}=\frac{a}{b}\), here \(a = 3\), \(b=1\).
\(\lim_{x
ightarrow0}\frac{\sin3x}{\sin x}=\lim_{x
ightarrow0}\frac{\sin3x}{3x}\cdot\frac{3x}{\sin x}\cdot3\)
Since \(\lim_{t
ightarrow0}\frac{\sin t}{t} = 1\), let \(t = 3x\) and \(t=x\) respectively.
\(\lim_{x
ightarrow0}\frac{\sin3x}{3x}\cdot\frac{3x}{\sin x}\cdot3=3\)

Step2: Direct substitution

For \(\lim_{x
ightarrow0}\frac{x}{x + 1}\), substitute \(x = 0\) directly.
\(\lim_{x
ightarrow0}\frac{x}{x + 1}=\frac{0}{0+1}=0\)

Step3: Use \(\lim_{t

ightarrow0}\frac{\sin t}{t}=1\)
For \(\lim_{x
ightarrow0^{+}}\frac{\sin4x^{2}}{x}\), rewrite it as \(\lim_{x
ightarrow0^{+}}\frac{\sin4x^{2}}{4x^{2}}\cdot4x\)
Let \(t = 4x^{2}\), when \(x
ightarrow0^{+}\), \(t
ightarrow0\).
\(\lim_{x
ightarrow0^{+}}\frac{\sin4x^{2}}{4x^{2}}\cdot4x=1\times0 = 0\)

Step4: L'Hopital's Rule

For \(\lim_{x
ightarrow\infty}\frac{x^{2}-1}{\ln x}\), it is in the \(\frac{\infty}{\infty}\) form.
Differentiate the numerator and denominator: \(\lim_{x
ightarrow\infty}\frac{2x}{\frac{1}{x}}=\lim_{x
ightarrow\infty}2x^{2}=\infty\)

Step5: L'Hopital's Rule (twice)

For \(\lim_{x
ightarrow\infty}\frac{x^{2}}{e^{x}}\), it is in the \(\frac{\infty}{\infty}\) form.
First - time differentiation: \(\lim_{x
ightarrow\infty}\frac{2x}{e^{x}}\) (still \(\frac{\infty}{\infty}\) form)
Second - time differentiation: \(\lim_{x
ightarrow\infty}\frac{2}{e^{x}}=0\)

Step6: Divide numerator and denominator by \(x^{2}\)

For \(\lim_{x
ightarrow\infty}\frac{x^{2}+2}{7x^{2}}=\lim_{x
ightarrow\infty}\frac{1+\frac{2}{x^{2}}}{7}\)
Since \(\lim_{x
ightarrow\infty}\frac{1}{x^{2}}=0\)
\(\lim_{x
ightarrow\infty}\frac{1+\frac{2}{x^{2}}}{7}=\frac{1 + 0}{7}=\frac{1}{7}\)

Answer:

(1) \(3\)
(2) \(0\)
(3) \(0\)
(4) \(\infty\)
(5) \(0\)
(6) \(\frac{1}{7}\)