A vertical asymptote occurs at \(\displaystyle x = a\) when \(\displaystyle \lim_{k \to a^+} f(x) = +/- \infty\)

or \(\displaystyle \lim_{k \to a^-} f(x) = +/- \infty\).

In our case, since we have a quotient of functions, we need only check for values of \(\displaystyle x\) that make the denominator \(\displaystyle 0\), but don't also make the numerator \(\displaystyle 0.\)

\(\displaystyle 0 = \ln(2)\cdot \sin(x)\)

This equals \(\displaystyle 0,\) when \(\displaystyle x\) is an integer multiple of \(\displaystyle \pi\).

Hence the vertical lines \(\displaystyle x = n\pi, n \in \mathbb{Z}\) are vertical asymptotes.

However we must exclude the case \(\displaystyle x = \pi\), because this will also cause the numerator to be \(\displaystyle 0\), thus creating a "hole" instead of an asymptote.

Hence our answer is

\(\displaystyle x = n\pi, n\in \mathbb{Z}, n \ne 1\).

We note that for all \(\displaystyle x\neq 0\), we have \(\displaystyle -1\le sin\frac{1}{x}\le 1\).

Hence,

\(\displaystyle 1\le {2+sin(\frac{1}{x})}\le 3\)

By inverting the above inequality and multiplying by x. We get the following:

\(\displaystyle \frac{x}{3}\le \frac{x}{2+sin(\frac{1}{x})}\le \frac{x}{2}\).

We know that,

\(\displaystyle \lim_{x\rightarrow 0}\frac{x}{2}=\lim_{x\rightarrow 0}\frac{x}{3}=0\)

and by the Squeeze Theorem,

we have:

\(\displaystyle \lim_{x\rightarrow 0}\frac{x}{2+sin(\frac{1}{x})}=0\)

We first need to see that the function sin(x) has infinitely many roots.

We can express these roots in the following form:

\(\displaystyle k\pi\), wkere k is an integer.

The function \(\displaystyle \frac{1}{sin(x)}\) has the roots as asymptotes.

Therefore this function's vertical asymptotes are expresses by \(\displaystyle k\pi\), where k is an integer. Since the integers are infinitely many, the vertical asymptotes are infinitely many.

To find the above limit, we need to note the following.

We have for all n positive integers:

\(\displaystyle x^n-1=(x-1)(x^{n-1}+x^{n-2}+\cdots+1)\).

(We can verify this formula by the long division)

Now we need to note that:

\(\displaystyle \frac{x^n-1}{x-1}=(x^{n-1}+x^{n-2}+\cdots+1)\), where \(\displaystyle x\neq 1\).

We have then:

\(\displaystyle \lim_{x\rightarrow 1}\frac{x^n-1}{x-1}=\lim_{x\rightarrow 1}(x^{n-1}+x^{n-2}+\cdots+1)\)

and we have

\(\displaystyle \lim_{x\rightarrow 1}(x^{n-1}+x^{n-2}+\cdots+1)=1+1...1(n times)\)\(\displaystyle =n\).

Since,

 \(\displaystyle \lim_{x\rightarrow 1}\frac{x^n-1}{x-1}=n\)

we obtain the following:

\(\displaystyle \lim_{x\rightarrow 1}\frac{x-1}{x^n-1}=\frac{1}{\lim_{x\rightarrow 1}\frac{x-1}{x^n-1}}=\frac{1}{n}\)

We need to notice that the function f is defined for all real numbers.

We need to also remark that for all reals:

\(\displaystyle x^2\ge 0\) implies that \(\displaystyle x^2+1\ge 1\)

this gives again:

\(\displaystyle \frac{1}{1+x^2}\le 1\) and therefore,

\(\displaystyle (\frac{1}{1+x^2})^n\le 1\).

This function can't be 0.

Assume for a moment that

\(\displaystyle (x^2+1)^n=0\), this implies that \(\displaystyle x^2=-1\) but this cannot happen since we are dealing with real numbers.

Therefore the above function can never be 0 and this means that it does not have a vertical asymptote. This is what we needed to show.

For this problem we first need to expand the denominator.

\(\displaystyle \lim_{x\rightarrow 2}\frac{x-2}{x^2-4}=\frac{x-2}{(x-2)(x+2)}\)

We can expand the denominator since \(\displaystyle x^2-4\) is a difference of squares.

From here we can cancel the \(\displaystyle (x-2)\) quantity from the numerator and denominator.

The resulting function is as follows:

\(\displaystyle \lim_{x\rightarrow2}\frac{1}{x+2}\)

Plugging in 2 we get our limit.

\(\displaystyle =\frac{1}{2+2}=\frac{1}{4}\)

We will use the following to prove this result.

Assuming that \(\displaystyle a,b\ge 0\)

\(\displaystyle (\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})=a-b\). We will use this result:

we have \(\displaystyle ({\sqrt{1+x}+\sqrt{1-x}})(\sqrt{1+x}-\sqrt{1-x})\)

\(\displaystyle =1+x-(1-x)=2x\)

Therefore

\(\displaystyle \lim_{x\rightarrow 0}{\frac{\sqrt{1+x}-\sqrt{1-x}}{x} =\)

\(\displaystyle \lim_{x\rightarrow 0}\frac{2x}{x({\sqrt{1+x}+\sqrt{1-x}})}\)

\(\displaystyle =\lim_{x\rightarrow 0}\frac{2}{({\sqrt{1+x}+\sqrt{1-x}})}\)

\(\displaystyle =\frac{2}{2}=1\)

this shows the limit is 1.

We will use the following identity to establish this result.

We have

\(\displaystyle a^3-b^3=(a-b)(a^2+ab+b^2)\)

and we note that :

\(\displaystyle \frac{\sqrt{3x}+3}{\sqrt{3x}+3}=1\)

Therefore by multiplying the above equivalency for 1 we get the following:

\(\displaystyle \frac{x^3-27}{\sqrt{3x}-3}=\frac{x^3-27}{\sqrt{3x}-3}\frac{\sqrt{3x+3}}{\sqrt{3x+3}}\)

 and we know that

\(\displaystyle (\sqrt{3x}-3)(\sqrt{3x}+3)=3x-9\)

\(\displaystyle \frac{x^3-27}{\sqrt{3x}-3}\frac{(\sqrt{3x}+3)}{(\sqrt{3x}+3)}= \frac{x^3-27}{\sqrt{3x}-3}=\frac{x^3-27}{3x-9}(\sqrt{3x}+3)= \frac{x^3-27}{3(x-3)}(\sqrt{3x}+3)\)

We can rewrite our equation using identities. 

\(\displaystyle x^3-27=x^3-3^3=(x-3)(x^2+3x+9)\)

This gives :

\(\displaystyle \frac{(x^2+3x+9)(\sqrt{3x}+3)}{3}\) and

now taking the limit as x goes to 3, we obtian

\(\displaystyle x=54\)

We note first that we can write:

\(\displaystyle x^7-1=(x-1)(x^6+x^5+x^4\cdots +1)\)

Therefore our expression becomes in this case:

\(\displaystyle \lim_{x\rightarrow 1}\frac{x^7-1}{1-x}=\lim_{x\rightarrow 1}\frac{(x^6+x^5\cdots+1)(x-1)}{1-x}\)

Noting now that:

\(\displaystyle \frac{x-1}{1-x}=-1\) for all \(\displaystyle x\neq 1\).

Therefore , we have: 

\(\displaystyle \lim_{x\rightarrow 1}\frac{x^7-1}{1-x}=\lim_{x\rightarrow 1}{-(x^6+x^5\cdots+1)}\)

and evaluating now for x=1, we obtain

\(\displaystyle \lim_{x\rightarrow 1}\frac{x^7-1}{1-x}=-7\)

We first note that the polynomial is defined for all real numbers.

We know that for any real number x different from 0, we have :

\(\displaystyle x^2\ge 0\).

Now we need to see that for any integer n we have:

\(\displaystyle x^{2n}\ge 0\). Adding in this case,

we have \(\displaystyle 1+x^2+x^4+\cdots+ x^{2n}> 1\)

and therefore \(\displaystyle p(x)>1\), this implies by definiton of q(x) that:

\(\displaystyle q(x)< 1\).

We also have \(\displaystyle q(x)>0\).

This means that  \(\displaystyle 0< q(x)< 1\).

Therefore q(x) can never be 0 and this means that it does not have an asymptote.