Differentiation (Part-38) | S N De

 

$\,6.\,$ Prove each of the following limits by intuitive and graphical methods :

$\,(i)\, \lim_{x \to -2} ~4=4.$

Sol. Clearly, here the given function say $\,f(x),\,$ is a constant function which is $\,f(x)=4.$ So, no matter in which way $\,x\,$ approaches $\,(-2),\,$ the limiting value of the guiven function is always the constant function itself which is $\,4.$

$\,(ii)\, \lim_{x \to 3}~~2x=6$

Sol. Suppose the variable $\,x\,$, always assuming positive values, increases without limit. Then we can prepare the following table showing the values of $\,x\,$ and $\,y\,$.

Let $\,x\,$ approaches $\,3\,$ from the right . We can take the values of $\,x$ as $\,3.01,3.001,3.001,\,\cdots$ etc. so that the corresponding values of $\,(2x)\,$ is $\,6.02,6.002,6.0002,\cdots$

Similarly, suppose the variable $\,x\,$, always assuming negative values, increases numerically without limit. Then we can prepare the following table showing the values of $\,x\,$ and $\,y\,$.

Let $\,x\,$ approaches $\,3\,$ from the left . We can take the values of $\,x$ as $\,2.99,2.999,2.9999\,\cdots$ etc. so that the corresponding values of $\,(2x)\,$ is $\,5.98,5.998,5.9998,\cdots$

Hence, from the aforementioned discussion, we can conclude that the limiting value of the given function $\,(2x),\,$ is $\,6.$

$\,(iii)\,\lim_{x \to 0}\frac{2x}{x}=2.$

Sol. Let $\,f(x)=\frac{2x}{x}.$

Now, since $\,x \to 0,\,~~ x \neq 0.$

So, $\,\,f(x)=\frac{2x}{x}=2\,\,$ and so $\,f(x)=2\,$ is a constant function. So, no matter in which way $\,x\,$ approaches $\,0,\,$ the limiting value of the guiven function is always the constant function itself which is $\,2.$

The graphical representation will be similar to solution of $\,6(i).$

$\,(iv)\,\lim_{x \to 0}~~|x|=0$

Sol. Let $\,f(x)=|x|.$

When $\,x \to 0-\,$ i.e., $\,x\,$ approaches $\,0\,$ from the left of the point $\,x=0,\,$ then we can have the following table showing the values of $\,x\,$ and $\,f(x).$

$\,x : \to ~~~~~~ -0.1,\,-0.01,-0.001,-0.0001,\cdots \\ f(x): \to ~~~~~~ 0.1,\,~~~~0.01,\,~~0.001,~~~0.0001,\cdots$

From the table, it is clear that when $\,x\,$ approaches $\,0\,$ from the left of the point  $\,x=0,\,\,f(x)\,$ gradually approaches $\,0,\,$ i.e., numerical difference between the value of $\,f(x)\,$ and $\,0\,$  [i.e., the value of $\,|f(x)-0|\,$, can be made as small as we like; in other words, $\,f(x) \to 0\,$ when $\,x \to  0-\,$ i.e., the left hand limit of $\,f(x)\,=\lim_{x \to 0-}~f(x)=0$ 

Again when $\,x \to 0+\,$ i.e., $\,x\,$ approaches $\,0\,$ from the right of the point $\,x=0,\,$ then we can have the following table showing the values of $\,x\,$ and $\,f(x).$

$\,x : \to ~~~~~~ 0.1,\,0.01,0.001,0.0001,\cdots \\ f(x): \to ~~ 0.1,\,0.01,\,0.001,0.0001,\cdots$

From the table, it is clear that when $\,x\,$ approaches $\,0\,$ from the right of the point  $\,x=0,\,\,f(x)\,$ gradually approaches $\,0,\,$ i.e., numerical difference between the value of $\,f(x)\,$ and $\,0\,$  [i.e., the value of $\,|f(x)-0|\,$, can be made as small as we like; in other words, $\,f(x) \to 0\,$ when $\,x \to  0+\,$ i.e., the right hand limit of $\,f(x)\,=\lim_{x \to 0+}~f(x)=0$

Hence, $\,\,\lim_{x \to 0-}~~|x|=\lim_{x \to 0+}~~|x|=0$

$\,(v)\,\,\lim_{x \to 3}\,\frac{x^2-9}{x-3}=6$

Sol.  Let $\,f(x)=\frac{x^2-9}{x-3}$

 To examine the existence of $~~\lim_{x \to 3}\,f(x)\,\,$ we prepare the following table showing the values of $\,x\,$ and $\,f(x)\,\,$ where the variable $\,x\,$ approaches $\,3\,$ either from the left or from the right of the point $\,x=3.$

$\,x : \to ~~~~~~ 2.9,\,2.99,2.999,2.9999,\cdots \\~f(x): \to ~ 5.9,\,5.99,\,5.999,5.9999,\cdots  $

$\,x : \to ~~~~~~3.1,3.01,3.001,\cdots \\~f(x): \to ~6.1, 6.01, 6.001, \cdots$ 

From the table it is clear that as $\,x\,$ gradually approaches $\,3\,$ [assuming values either less than or greater than $\,3\,$ and sufficiently close $\,3\,$]  the values of $\,f(x)\,$ gradually approaches $\,6\,$ i.e., the numerical difference between the values of $\,f(x)\,$ and $\,6\,$ ( or, the value of $\,|f(x)-6|$ ) can be made less than any pre-assigned positive number however small . In other words, $\,f(x) \to 6,\,$ when $\,x \to 3.$ 

Therefore $~~\lim_{x \to 3}\,f(x)\,\,$ exists and its value is $\,6.$ 

$\,7.\,~~f(x)=\sqrt{x-3},\,$ does $\,\lim_{x \to 3}~~f(x)\,$ exist ? Give reasons.

Sol.  We know , $\,\lim_{x \to a}~~f(x)\,\,$ exists , if 

$\,(i)\,\,\lim_{x \to a+}~~f(x)\,$ and $\lim_{x \to a-}~~f(x)\,$ exists and 

$\,(ii)~~\lim_{x \to a-}~~f(x)=\lim_{x \to a+}~~f(x).$

Now, for $\,x \to 3+,\,\,$ suppose that $\,\,x=3+h,\,(h>0)[\because x \to 3+ \Rightarrow x>3]$

So, in this case, $\,f(3+h)=\sqrt{3+h-3}=\sqrt h \rightarrow(1)$

Again, for $\,x \to 3-,\,\,$ suppose that $\,\,x=3-h,\,(h>0)[\because x \to 3- \Rightarrow x<3]$

So, in this case, $\,f(3-h)=\sqrt{3-h-3}=\sqrt{-h} \rightarrow(2)$

From $\,(2),\,$ we notice that left hand limit of the given function does not exist as it gives imaginary value.

Also, $\,f(3+h) \neq f(3-h)$ 

So, from this discussion, we can conclude that $\,\lim_{x \to 3}~~f(x)\,$ does not exist.

Alternatively, we can also follow S N De workout example no. $\,2\,$ to solve this problem differently.

$\,8.\,$ A function $\,\phi(x)\,\,$ is defined as follows :

$\,\phi(x)=2,\,\,\text{when}\,\, x \neq 1\\~~~~~~~~~=0,\,\,\text{when}\,\, x =1.$

Does  $\,\lim_{x \to 1}\,\phi(x)\,$ exist? If so, find its value.

Sol.  We know , $\,\lim_{x \to a}~~f(x)\,\,$ exists , if 

$\,(i)\,\,\lim_{x \to a+}~~f(x)\,$ and $\lim_{x \to a-}~~f(x)\,$ exists and 

$\,(ii)~~\lim_{x \to a-}~~f(x)=\lim_{x \to a+}~~f(x).$

From the definition of the function we notice that, both  $\,\lim_{x \to 1+}\,\phi(x)\,$ and  $\,\lim_{x \to 1-}\,\phi(x)\,$ exist and 

 $\,\lim_{x \to 1+}\,\phi(x)=2=\lim_{x \to 1-}\,\phi(x)\\~~[\because x \to 1+ ~\text{or,}~~x \to 1- \Rightarrow x \neq 1]$

Hence, $\,\lim_{x \to 1}\,\phi(x)\,$ exists and its value is $\,2.$

$\,9.\,$ If $\,f(x)=\frac{|x|}{x},\,$ show that, $\,\lim_{x \to 0}~~f(x)\,\,$ does not exist.

Sol. We know, $\,|x|=x,\, \text{if}\,\, x \geq 0\\~~~~~=-x,\, \text{if}\,\, x <0$

Now, $\,\lim_{x \to 0-}\,f(x)\\=\lim_{x \to 0-}\frac{|x|}{x}\\=\lim_{x \to 0-}\left(\frac{-x}{x}\right)\,\,[\because x \to 0-,\,x<0]\\=-1 \rightarrow(1)$

Also, $\,\lim_{x \to 0+}\,f(x)\\=\lim_{x \to 0+}\frac{|x|}{x}\\=\lim_{x \to 0+}\left(\frac{x}{x}\right)\,\,[\because x \to 0+,\,x \geq 0]\\=1\rightarrow(2)$

 So, from $\,(1)\,$ and $\,(2),\,$ we can say

$\,\lim_{x \to 0+}\,f(x) \neq \lim_{x \to 0-}\,f(x) \rightarrow(3)$

Hence,  from $\,(3),\,$ we can say that $\,\lim_{x \to 0}~~f(x)\,\,$ does not exi

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