Sandwich theorem Questions in English

(B) Given $f(x) = x\sin \frac{1}{x}$ for $x \ne 0$ and $f(0) = 0$.
We know that for all $x \ne 0$,$-1 \le \sin \frac{1}{x} \le 1$.
Multiplying by $|x|$,we get $-|x| \le x\sin \frac{1}{x} \le |x|$.
Applying the Squeeze Theorem,since $\lim_{x \to 0} (-|x|) = 0$ and $\lim_{x \to 0} |x| = 0$,it follows that $\lim_{x \to 0} x\sin \frac{1}{x} = 0$.
Thus,$\lim_{x \to 0} f(x) = 0$.

(B) We need to evaluate $\mathop {\lim }\limits_{x \to \infty } \frac{{\sin x}}{x}$.
We know that for all $x \in \mathbb{R}$,the value of $\sin x$ is bounded such that $-1 \le \sin x \le 1$.
For $x > 0$,we can divide the inequality by $x$:
$-\frac{1}{x} \le \frac{\sin x}{x} \le \frac{1}{x}$.
Applying the Squeeze Theorem as $x \to \infty$:
$\mathop {\lim }\limits_{x \to \infty } \left( -\frac{1}{x} \right) = 0$ and $\mathop {\lim }\limits_{x \to \infty } \left( \frac{1}{x} \right) = 0$.
Therefore,by the Squeeze Theorem,$\mathop {\lim }\limits_{x \to \infty } \frac{{\sin x}}{x} = 0$.

(A) For any sequence $x_n$ such that $x_n \to 0$,we have $|f(x_n)| = |x_n|$.
Since $\lim_{n \to \infty} |x_n| = 0$,by the Squeeze Theorem,$\lim_{n \to \infty} f(x_n) = 0$.
Therefore,$\lim_{x \to 0} f(x) = 0$.

(D) Since $f(x)$ is a positive increasing function,for $x > 0$,we have $f(x) < f(2x) < f(3x)$.
Dividing by $f(x) > 0$,we get $1 < \frac{f(2x)}{f(x)} < \frac{f(3x)}{f(x)}$.
Taking the limit as $x \to \infty$,we have $\lim_{x \to \infty} 1 \le \lim_{x \to \infty} \frac{f(2x)}{f(x)} \le \lim_{x \to \infty} \frac{f(3x)}{f(x)}$.
Given that $\lim_{x \to \infty} \frac{f(3x)}{f(x)} = 1$,by the Sandwich theorem,we have $1 \le \lim_{x \to \infty} \frac{f(2x)}{f(x)} \le 1$.
Therefore,$\lim_{x \to \infty} \frac{f(2x)}{f(x)} = 1$.

(B) We are given the inequality: $\frac{\cos(2x - 4) - 33}{2} < f(x) < \frac{x^2 |4x - 8|}{x - 2}$.
First,evaluate the limit of the left-hand side as $x \to 2^-$:
$\lim_{x \to 2^-} \frac{\cos(2x - 4) - 33}{2} = \frac{\cos(0) - 33}{2} = \frac{1 - 33}{2} = \frac{-32}{2} = -16$.
Next,evaluate the limit of the right-hand side as $x \to 2^-$:
Since $x < 2$,$x - 2 < 0$,so $|4x - 8| = |4(x - 2)| = -4(x - 2)$.
$\lim_{x \to 2^-} \frac{x^2 |4x - 8|}{x - 2} = \lim_{x \to 2^-} \frac{x^2 (-4(x - 2))}{x - 2} = \lim_{x \to 2^-} (-4x^2) = -4(2^2) = -16$.
By the Sandwich Theorem,since both the left and right limits are $-16$,we have $\lim_{x \to 2^-} f(x) = -16$.

(C) Given $1 < f(x) \cdot g(x) < 10$ and $g(x) = x + \frac{1}{x} = \frac{x^2 + 1}{x}$.
Substituting $g(x)$,we get $1 < f(x) \cdot \left( \frac{x^2 + 1}{x} \right) < 10$.
Multiplying by $\frac{x}{x^2 + 1}$ (which is positive for $x > 0$),we get $\frac{x}{x^2 + 1} < f(x) < \frac{10x}{x^2 + 1}$.
Applying the limit as $x \to \infty$ to all parts:
$\lim_{x \to \infty} \frac{x}{x^2 + 1} = 0$ and $\lim_{x \to \infty} \frac{10x}{x^2 + 1} = 0$.
By the Sandwich Theorem,$\lim_{x \to \infty} f(x) = 0$.

(B) Let $L = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{x - \sin x}}{x}} \right)\sin \left( {\frac{1}{x}} \right)$.
We can rewrite the expression as:
$L = \mathop {\lim }\limits_{x \to 0} \left( {1 - \frac{\sin x}{x}} \right) \sin \left( {\frac{1}{x}} \right)$.
We know that $\mathop {\lim }\limits_{x \to 0} \frac{\sin x}{x} = 1$,so $\mathop {\lim }\limits_{x \to 0} \left( {1 - \frac{\sin x}{x}} \right) = 1 - 1 = 0$.
The term $\sin \left( {\frac{1}{x}} \right)$ oscillates between $-1$ and $1$ as $x \to 0$,meaning it is a bounded function.
By the Squeeze Theorem,since the limit of the first part is $0$ and the second part is bounded,the product is $0 \times (\text{bounded value}) = 0$.
Therefore,the limit is $0$.

(D) Given $\frac{x}{[x]} \le f(x) \le \sqrt{6 - x}$ for $x \ne 2$.
For $x \to 2^-$,$[x] = 1$. Thus,$\lim_{x \to 2^-} \frac{x}{[x]} = \frac{2}{1} = 2$ and $\lim_{x \to 2^-} \sqrt{6 - x} = \sqrt{6 - 2} = 2$.
By the Sandwich Theorem,$\lim_{x \to 2^-} f(x) = 2$. So,Statement $1$ is true.
For $x \to 2^+$,$[x] = 2$. Thus,$\lim_{x \to 2^+} \frac{x}{[x]} = \frac{2}{2} = 1$ and $\lim_{x \to 2^+} \sqrt{6 - x} = 2$.
By the Sandwich Theorem,$1 \le \lim_{x \to 2^+} f(x) \le 2$. Since the left-hand limit is $2$ and the right-hand limit is not necessarily $2$,the limit $\lim_{x \to 2} f(x)$ does not exist.
Since $\lim_{x \to 2} f(x)$ does not exist,$f$ is not continuous at $x = 2$. So,Statement $2$ is false.

(A) We know that for any real number $x$,$x-1 < [x] \leq x$.
Applying this to the terms $[kr]$ for $k=1, 2, \ldots, n$,we have:
$kr-1 < [kr] \leq kr$.
Summing these inequalities from $k=1$ to $n$:
$\sum_{k=1}^{n} (kr-1) < \sum_{k=1}^{n} [kr] \leq \sum_{k=1}^{n} kr$.
$r \frac{n(n+1)}{2} - n < \sum_{k=1}^{n} [kr] \leq r \frac{n(n+1)}{2}$.
Dividing the entire inequality by $n^2$:
$\frac{r \frac{n(n+1)}{2} - n}{n^2} < \frac{\sum_{k=1}^{n} [kr]}{n^2} \leq \frac{r \frac{n(n+1)}{2}}{n^2}$.
Taking the limit as $n \rightarrow \infty$:
$\lim_{n \rightarrow \infty} \left( \frac{r(n^2+n)}{2n^2} - \frac{n}{n^2} \right) = \frac{r}{2}$
and
$\lim_{n \rightarrow \infty} \frac{r(n^2+n)}{2n^2} = \frac{r}{2}$.
By the Sandwich Theorem,the limit is $\frac{r}{2}$.

(C) Given the sequence $s_n = \sum_{k=0}^n \frac{1}{\sqrt{n^2+k}}$.
For each $k$ such that $0 \leq k \leq n$,we have $n^2 \leq n^2+k \leq n^2+n$.
Taking the square root and reciprocal,we get $\frac{1}{\sqrt{n^2+n}} \leq \frac{1}{\sqrt{n^2+k}} \leq \frac{1}{\sqrt{n^2}}$.
Summing from $k=0$ to $n$ (which contains $n+1$ terms),we have:
$(n+1) \frac{1}{\sqrt{n^2+n}} \leq s_n \leq (n+1) \frac{1}{\sqrt{n^2}}$.
As $n \rightarrow \infty$,the left side $\frac{n+1}{\sqrt{n^2+n}} = \frac{n(1+1/n)}{n\sqrt{1+1/n}} = \sqrt{1+1/n} \rightarrow 1$.
The right side $\frac{n+1}{n} = 1 + \frac{1}{n} \rightarrow 1$.
By the Squeeze Theorem,$\lim_{n \rightarrow \infty} s_n = 1$.
Since $1$ lies in the interval $[1, 2)$,the correct option is $C$.

(D) Let $L_n = \prod_{k=1}^{n} \left(2^{\frac{1}{2}} - 2^{\frac{1}{2k+1}}\right)$.
Note that for all $k \ge 1$,$2^{\frac{1}{2}} - 2^{\frac{1}{2k+1}} > 0$.
Also,$2^{\frac{1}{2}} - 2^{\frac{1}{2k+1}} < 2^{\frac{1}{2}} - 2^0 = \sqrt{2} - 1 < 1$.
Since $0 < 2^{\frac{1}{2}} - 2^{\frac{1}{2k+1}} < \sqrt{2} - 1 < 1$ for all $k$,the product $L_n$ is a product of $n$ terms,each of which is less than $k_0 = \sqrt{2} - 1 < 1$.
Thus,$0 < L_n < (\sqrt{2} - 1)^n$.
As $n \rightarrow \infty$,$(\sqrt{2} - 1)^n \rightarrow 0$ because $|\sqrt{2} - 1| < 1$.
By the Squeeze Theorem,$\lim_{n \rightarrow \infty} L_n = 0$.

(B) Given that $f: R \rightarrow (0, \infty)$ is a strictly increasing function.
We are given the limit $\lim _{x \rightarrow \infty} \frac{f(7 x)}{f(x)}=1$.
Since $f$ is a strictly increasing function and $x < 5x < 7x$ for $x > 0$,we have $f(x) < f(5x) < f(7x)$.
Dividing the inequality by $f(x) > 0$,we get $1 < \frac{f(5x)}{f(x)} < \frac{f(7x)}{f(x)}$.
Taking the limit as $x \rightarrow \infty$,we have $\lim _{x \rightarrow \infty} 1 \leq \lim _{x \rightarrow \infty} \frac{f(5x)}{f(x)} \leq \lim _{x \rightarrow \infty} \frac{f(7x)}{f(x)}$.
Substituting the given limit,we get $1 \leq \lim _{x \rightarrow \infty} \frac{f(5x)}{f(x)} \leq 1$.
By the Squeeze Theorem,$\lim _{x \rightarrow \infty} \frac{f(5x)}{f(x)} = 1$.
Therefore,the value of $\lim _{x \rightarrow \infty} \left[\frac{f(5x)}{f(x)} - 1\right] = 1 - 1 = 0$.

(B) We need to evaluate the limit $L = \lim _{x \rightarrow 0} x^7 \left[ \frac{1}{x^3} \right]$.
Using the property of the greatest integer function,$\frac{1}{x^3} - 1 < \left[ \frac{1}{x^3} \right] \le \frac{1}{x^3}$.
Case $1$: $x > 0$. Multiplying by $x^7$ (which is positive),we get $x^7 \left( \frac{1}{x^3} - 1 \right) < x^7 \left[ \frac{1}{x^3} \right] \le x^7 \left( \frac{1}{x^3} \right)$.
$x^4 - x^7 < x^7 \left[ \frac{1}{x^3} \right] \le x^4$.
As $x \rightarrow 0^+$,both $x^4 - x^7 \rightarrow 0$ and $x^4 \rightarrow 0$. By the Squeeze Theorem,the limit is $0$.
Case $2$: $x < 0$. Multiplying by $x^7$ (which is negative),the inequality signs reverse: $x^7 \left( \frac{1}{x^3} \right) \le x^7 \left[ \frac{1}{x^3} \right] < x^7 \left( \frac{1}{x^3} - 1 \right)$.
$x^4 \le x^7 \left[ \frac{1}{x^3} \right] < x^4 - x^7$.
As $x \rightarrow 0^-$,both $x^4 \rightarrow 0$ and $x^4 - x^7 \rightarrow 0$. By the Squeeze Theorem,the limit is $0$.
Since the left-hand limit and right-hand limit are both $0$,the limit is $0$.

(A) Given,$\lim _{x \rightarrow 0} \frac{|x|}{\sqrt{x^4+4 x^2+5}}=k$.
Substituting $x=0$,we get $k = \frac{0}{\sqrt{0+0+5}} = 0$.
Now,consider $\lim _{x \rightarrow 0} x^4 \sin \left(\frac{1}{3 \sqrt{x}}\right)=l$.
Since $\sin \left(\frac{1}{3 \sqrt{x}}\right)$ oscillates between $-1$ and $1$ as $x \rightarrow 0$,and $x^4 \rightarrow 0$,by the Squeeze Theorem,$0 \times (\text{finite value}) = 0$.
Thus,$l = 0$.
Therefore,$k+l = 0+0 = 0$.

(A) We know that for any real number $x$,the value of $\sin x$ lies in the interval $[-1, 1]$.
Therefore,$2 + \sin x$ lies in the interval $[2 - 1, 2 + 1] = [1, 3]$.
As $x \rightarrow \infty$,the denominator $x^2 + 3 \rightarrow \infty$.
Thus,we have $\lim _{x}$ ${\rightarrow \infty}\left(\frac{2+\sin x}{x^2+3}\right) = \frac{\text{finite value between } 1 \text{ and } 3}{\infty} = 0$.

(D) We know that for any real number $y$,$y - 1 < [y] \leq y$.
Substituting $y = 2x - 3$,we get:
$(2x - 3) - 1 < [2x - 3] \leq 2x - 3$
$2x - 4 < [2x - 3] \leq 2x - 3$
Dividing the entire inequality by $x$ (for $x > 0$):
$\frac{2x - 4}{x} < \frac{[2x - 3]}{x} \leq \frac{2x - 3}{x}$
$2 - \frac{4}{x} < \frac{[2x - 3]}{x} \leq 2 - \frac{3}{x}$
Applying the limit as $x \rightarrow \infty$:
$\lim_{x \rightarrow \infty} (2 - \frac{4}{x}) \leq \lim_{x \rightarrow \infty} \frac{[2x - 3]}{x} \leq \lim_{x \rightarrow \infty} (2 - \frac{3}{x})$
$2 - 0 \leq \lim_{x \rightarrow \infty} \frac{[2x - 3]}{x} \leq 2 - 0$
By the Squeeze Theorem,$\lim_{x \rightarrow \infty} \frac{[2x - 3]}{x} = 2$.

(D) We evaluate the limit by splitting it into two parts:
$\lim _{x \rightarrow 2} \left[ (x-2)^2 \cos \left(\frac{2}{x-2}\right) \right] + \lim _{x \rightarrow 2} \left[ \frac{x^2-4}{x^3-2 x-4} \right]$
For the first part,$\lim _{x \rightarrow 2} (x-2)^2 \cos \left(\frac{2}{x-2}\right)$. Since $(x-2)^2 \rightarrow 0$ and $\cos \left(\frac{2}{x-2}\right)$ is bounded in $[-1, 1]$,by the Squeeze Theorem,the limit is $0 \times [-1, 1] = 0$.
For the second part,$\lim _{x \rightarrow 2} \frac{(x-2)(x+2)}{(x-2)(x^2+2x+2)} = \lim _{x \rightarrow 2} \frac{x+2}{x^2+2x+2}$.
Substituting $x=2$,we get $\frac{2+2}{2^2+2(2)+2} = \frac{4}{4+4+2} = \frac{4}{10} = \frac{2}{5}$.
Thus,the total limit is $0 + \frac{2}{5} = \frac{2}{5}$.

(B) We use the Squeeze Theorem to evaluate the limit.
We know that for all $x \neq 0$,$-1 \leq \sin \left(\frac{\pi}{x}\right) \leq 1$.
Multiplying the inequality by $x^2$ (since $x^2 > 0$ for $x \neq 0$),we get:
$-x^2 \leq x^2 \sin \left(\frac{\pi}{x}\right) \leq x^2$.
Now,taking the limit as $x \rightarrow 0$ on all sides:
$\lim _{x \rightarrow 0} (-x^2) \leq \lim _{x \rightarrow 0} x^2 \sin \left(\frac{\pi}{x}\right) \leq \lim _{x \rightarrow 0} x^2$.
Since $\lim _{x \rightarrow 0} (-x^2) = 0$ and $\lim _{x \rightarrow 0} x^2 = 0$,by the Squeeze Theorem,$\lim _{x \rightarrow 0} x^2 \sin \left(\frac{\pi}{x}\right) = 0$.

(A) Given that $f$ is an increasing function and $f(x) > 0$ for all $x \in R^{+}$.
We are given that $\lim _{x \rightarrow \infty} \frac{f(9 x)}{f(3 x)}=1$.
Since $f$ is an increasing function,for any $x > 0$,we have the inequality:
$3x < 6x < 9x$
$\Rightarrow f(3x) < f(6x) < f(9x)$
Dividing the entire inequality by $f(3x)$ (which is positive),we get:
$\frac{f(3x)}{f(3x)} < \frac{f(6x)}{f(3x)} < \frac{f(9x)}{f(3x)}$
$1 < \frac{f(6x)}{f(3x)} < \frac{f(9x)}{f(3x)}$
Now,taking the limit as $x \rightarrow \infty$ on all sides:
$\lim _{x \rightarrow \infty} 1 \leq \lim _{x \rightarrow \infty} \frac{f(6x)}{f(3x)} \leq \lim _{x \rightarrow \infty} \frac{f(9x)}{f(3x)}$
$1 \leq \lim _{x \rightarrow \infty} \frac{f(6x)}{f(3x)} \leq 1$
By the Sandwich Theorem (or Squeeze Theorem),we conclude that:
$\lim _{x \rightarrow \infty} \frac{f(6x)}{f(3x)} = 1$.

(A) We know that for any real number $y$,$y - 1 < [y] \leq y$.
Substituting $y = \frac{q}{x}$,we get $\frac{q}{x} - 1 < \left[ \frac{q}{x} \right] \leq \frac{q}{x}$.
Multiplying by $\frac{x}{p}$ (assuming $x > 0$ and $p > 0$),we get $\frac{x}{p} \left( \frac{q}{x} - 1 \right) < \frac{x}{p} \left[ \frac{q}{x} \right] \leq \frac{x}{p} \left( \frac{q}{x} \right)$.
This simplifies to $\frac{q}{p} - \frac{x}{p} < \frac{x}{p} \left[ \frac{q}{x} \right] \leq \frac{q}{p}$.
Taking the limit as $x \rightarrow 0^{+}$,by the Squeeze Theorem,the value is $\frac{q}{p}$.

(C) We know that for any real number $x$,$x-1 < [x] \leq x$.
Substituting $x = n\sqrt{2}$,we get $n\sqrt{2}-1 < [n\sqrt{2}] \leq n\sqrt{2}$.
Dividing the entire inequality by $n$ (where $n > 0$),we get $\frac{n\sqrt{2}-1}{n} < \frac{[n\sqrt{2}]}{n} \leq \frac{n\sqrt{2}}{n}$.
This simplifies to $\sqrt{2} - \frac{1}{n} < \frac{[n\sqrt{2}]}{n} \leq \sqrt{2}$.
Taking the limit as $n \rightarrow \infty$,we have $\lim_{n \rightarrow \infty} (\sqrt{2} - \frac{1}{n}) = \sqrt{2}$ and $\lim_{n \rightarrow \infty} \sqrt{2} = \sqrt{2}$.
By the Sandwich Theorem,$\lim_{n \rightarrow \infty} \frac{[n\sqrt{2}]}{n} = \sqrt{2}$.

(A) We need to evaluate $\lim_{x \rightarrow 0} x \sin \left(e^{\frac{1}{x}}\right)$.
Since $-1 \leq \sin \theta \leq 1$ for any real $\theta$,we have $-1 \leq \sin \left(e^{\frac{1}{x}}\right) \leq 1$.
Multiplying by $x$ (assuming $x > 0$): $-x \leq x \sin \left(e^{\frac{1}{x}}\right) \leq x$.
As $x \rightarrow 0^+$,both $-x$ and $x$ approach $0$. By the Squeeze Theorem,$\lim_{x \rightarrow 0^+} x \sin \left(e^{\frac{1}{x}}\right) = 0$.
For $x < 0$,let $x = -h$ where $h > 0$. As $x \rightarrow 0^-$,$h \rightarrow 0^+$.
The expression becomes $\lim_{h \rightarrow 0^+} (-h) \sin \left(e^{-\frac{1}{h}}\right)$.
As $h \rightarrow 0^+$,$e^{-\frac{1}{h}} \rightarrow e^{-\infty} = 0$,so $\sin \left(e^{-\frac{1}{h}}\right) \rightarrow \sin(0) = 0$.
Thus,$\lim_{h \rightarrow 0^+} (-h) \cdot 0 = 0$.
Since the left-hand limit and right-hand limit are both $0$,the limit exists and is equal to $0$.