Question

show that lim x tends to 2- x/[x] is not equal to lim x tends to 2+ x/[x]​

Answer 1

Given Question :-

Show that,

$\displaystyle\lim_{x \to 2^-}\rm \frac{x}{[x]} \: \ne \: \: \displaystyle\lim_{x \to 2^ + }\rm \frac{x}{[x]}$

where, [ . ] denotes greatest integer function.

$\large\underline{\sf{Solution-}}$

Consider,

$\rm \: \displaystyle\lim_{x \to 2^-}\rm \frac{x}{[x]}$

To evaluate this limit, we use method of Substitution

So, Substitute

$\rm \: x = 2 - h, \: \: as \: x \: \to \: 2, \: \: so \: h \: \to \: 0$

So, on substituting the values, we get

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm \frac{2 - h}{[2 - h]}$

Now, By definition of Greatest Integer function,

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf [x] \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 1 \leqslant x < 2 & \sf 1 \\ \\ \sf 2 \leqslant x < 3 & \sf 2 \end{array}} \\ \end{gathered}$

So, using this definition, we have

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm \frac{2 - h}{1}$

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm (2 - h)$

$\rm \: = \: 2$

$\rm\implies \: \boxed{\sf{ \: \:\displaystyle\lim_{x \to 2^-}\rm \frac{x}{[x]} = 2 \: }} \: - - - (1)$

Now, Consider

$\rm \: \displaystyle\lim_{x \to 2^ + }\rm \frac{x}{[x]}$

Now, to evaluate this limit, we have to use method of Substitution

So, Substitute

$\rm \: x = 2 + h, \: \: as \: x \: \to \: 2, \: \: so \: h \: \to \: 0$

So, on substituting these values, we get

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm \frac{2 + h}{[2 + h]}$

So, by definition of Greatest Integer function, we have

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf [x] \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 1 \leqslant x < 2 & \sf 1 \\ \\ \sf 2 \leqslant x < 3 & \sf 2 \end{array}} \\ \end{gathered}$

So, using this definition, we have

$\rm \: = \: \displaystyle\lim_{h \to 0}\rm \frac{2 + h}{2}$

$\rm \: = \: \dfrac{2}{2}$

$\rm \: = \: 1$

So,

$\rm\implies \: \boxed{\sf{ \: \:\displaystyle\lim_{x \to 2^ + }\rm \frac{x}{[x]} = 1 \: }} \: - - - (2)$

So, from equation (1) and (2), we concluded that

$\: \: \: \: \: \: \: \: \quad\boxed{\sf{ \: \: \displaystyle\lim_{x \to 2^-}\rm \frac{x}{[x]} \: \ne \: \: \displaystyle\lim_{x \to 2^ + }\rm \frac{x}{[x]} \: \: \: }}$

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Additional Information

$\boxed{\sf{ \:\displaystyle\lim_{x \to 0}\rm \frac{sinx}{x} \: = \: 1 \: }}$

$\boxed{\sf{ \:\displaystyle\lim_{x \to 0}\rm \frac{tanx}{x} \: = \: 1 \: }}$

$\boxed{\sf{ \:\displaystyle\lim_{x \to 0}\rm \frac{ {e}^{x} - 1}{x} \: = \: 1 \: }}$

$\boxed{\sf{ \:\displaystyle\lim_{x \to 0}\rm \frac{ {a}^{x} - 1}{x} \: = \: loga \: }}$

$\boxed{\sf{ \:\displaystyle\lim_{x \to 0}\rm \frac{ log(1 + x)}{x} \: = \: 1 \: }}$

Answer 2

Step-by-step explanation:

$\tt\[ \underset{x \rightarrow 2^{-}} { \lim}\dfrac{x}{[x]} \]$

Let $\sf x=2-h$, where $\sf h \rightarrow 0$

$\tt \: \quad \quad \qquad \lim _{h \rightarrow 0} \dfrac{(2-h)}{[2-h]} =\dfrac{2}{1} \\ \tt \lim _{x \rightarrow 2^{+}} \dfrac{x}{[x]}$

Let $\sf x=2+h$, where $\sf h \rightarrow 0$

$\[ \begin{array}{l} \underset{ \tt \: h \rightarrow 0}{ \lim} \tt\dfrac{(2+h)}{[2+h]} =\dfrac{2}{2} =1 \end{array} \]$

$\tt \: Thus, \underset{x \rightarrow 2^{-}}{ \lim}\dfrac{x}{[x]} \neq \underset{x \longrightarrow 2^{+}}{ \lim} \dfrac{x}{[x]}$