Question

find the limit or show it does not exist \\( \lim_{x \to \infty} x^{1/3} \sin\left( \frac{1}{\sqrt{x}} \
ight) \\). find the limit (you can use lhospitals rule) \\( \lim_{x \to 0} \frac{\ln(1 + \ln(1 + 3x^2))}{x^2} \\).

Explanation:

Step1: Check the form of the limit

We need to find $\lim_{x \to 0} \frac{\ln(1 + \ln(1 + 3x^2))}{x^2}$. First, substitute $x = 0$ into the numerator and the denominator.

• Numerator: $\ln(1 + \ln(1 + 3 \cdot 0^2))=\ln(1 + \ln(1))=\ln(1 + 0)=0$
• Denominator: $0^2 = 0$

So the limit is in the $\frac{0}{0}$ indeterminate form, and we can use L'Hospital's Rule.

Step2: Apply L'Hospital's Rule (first time)

L'Hospital's Rule states that if $\lim_{x \to a} \frac{f(x)}{g(x)}$ is $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim_{x \to a} \frac{f(x)}{g(x)}=\lim_{x \to a} \frac{f'(x)}{g'(x)}$.

Let $f(x)=\ln(1 + \ln(1 + 3x^2))$ and $g(x)=x^2$.

Find $f'(x)$ using the chain rule:

• Let $u = 1 + \ln(1 + 3x^2)$, then $f(u)=\ln(u)$, so $f'(u)=\frac{1}{u}$.
• Now find $u'$: let $v = 1 + 3x^2$, then $u = 1 + \ln(v)$, so $u'=\frac{1}{v}\cdot v'$.
• $v' = 6x$.

Putting it all together: $u'=\frac{1}{1 + 3x^2}\cdot 6x$, and $f'(x)=\frac{1}{1 + \ln(1 + 3x^2)}\cdot\frac{6x}{1 + 3x^2}$

$g'(x)=2x$

So now we have $\lim_{x \to 0} \frac{\frac{6x}{(1 + \ln(1 + 3x^2))(1 + 3x^2)}}{2x}$

Simplify the expression: $\lim_{x \to 0} \frac{6x}{2x(1 + \ln(1 + 3x^2))(1 + 3x^2)}=\lim_{x \to 0} \frac{3}{(1 + \ln(1 + 3x^2))(1 + 3x^2)}$ (we can cancel $x$ since $x \to 0$ but $x
eq0$)

Step3: Evaluate the new limit

Substitute $x = 0$ into the new expression:

• Denominator: $(1 + \ln(1 + 3 \cdot 0^2))(1 + 3 \cdot 0^2)=(1 + \ln(1))(1)= (1 + 0)\cdot1 = 1$
• Numerator: $3$

So $\lim_{x \to 0} \frac{3}{(1 + \ln(1 + 3x^2))(1 + 3x^2)}=\frac{3}{1}=3$

Answer:

$\boxed{3}$