Differentiation (Part-38) | S N De

 $\,1.\,$ The function $\,f(x)\,$ is defined as follows :

$f(x)=2-3x,\,\,\text{when}\,\, x<0 \\~~~~~~~=3x-2,\,\,\text{when}\,\, x \geq 0 $

Evaluate $\,\lim_{x \to 0+}f(x)\,\,\text{and}\,\,\lim_{x \to 0-}f(x)\,$ and hence state whether $\,\lim_{x \to 0}f(x)\,\,$ exist or not.

Sol. $\, \lim_{x \to 0+}f(x)\\=\lim_{x \to 0+}(3x-2)~~~[ \because x \to 0+ ,~~x >0]\\=3 \times 0-2\\=-2$

$\, \lim_{x \to 0-}f(x)\\=\lim_{x \to 0-}(2-3x)~~~[ \because x \to 0- ,~~x <0]\\=2-3 \times 0\\=2$

$\therefore \lim_{x \to 0+}f(x) \neq \lim_{x \to 0-}f(x)$

Hence, $\,\lim_{x \to 0}f(x)\,\,$ does not exist.

$\,2.\,$ If $\,\phi(x)=e^{2x}+1,~~\text{when}~~x \geq 0\\~~~~~~~~~=1+\cos 2x,~~\text{when}~~x <0$

show that , $\,\lim_{x \to 0}\phi(x)=2.$

Sol. $\,\lim_{x \to 0+}\phi(x)\\=\lim_{x \to 0+}[e^{2x}+1]\\=e^{2 \times 0}+1\\=1+1=2.\\ \lim_{x \to 0-} \phi(x)\\=\lim_{x \to 0-}[(1+\cos2x)]\\=1+\cos(2 \times 0)\\=1+1\\=2\\ \therefore \lim_{x \to 0+}\phi(x)=\lim_{x \to 0-}\phi(x)=2 \\ \text{Hence},\,\, \phi(x)=2\,\,\text{(showed)}$

$\,3.\,$ If $\,f(x)=[x],\,$ where  $\,[x]\,$ denotes greatest integer less than or equal to $\,x,\,$ show that, $\,\lim_{x \to 2} \,f(x)\,$ does not exist.

Sol.  $\,[x]=1,\,\text{when}\,1\leq x < 2\\~~~~~= 2,\,\text{when}\,2\leq x < 3$

So, $\,\lim_{x \to 2+}f(x)=\lim_{x \to 2+}[x]=2\\\lim_{x \to 2-}f(x)=\lim_{x \to 2-}[x]=1\\ \therefore \lim_{x \to 2+}f(x) \neq \lim_{x \to 2-}f(x)$

Hence,  $\,\lim_{x \to 2} \,f(x)\,$ does not exist.

$\,4.\,$ Evaluate : $\,\lim_{x \to 1}\,\frac{x^2-3x+2}{x^3-4x+3}$

Sol. At first, we factorize $\,x^2-3x+2\,$ and $\,x^3-4x+3$.

$\,~~~~x^2-3x+2\\=x^2-2x-x+2\\=x(x-2)-1(x-2)\\=(x-2)(x-1)\rightarrow(1)$

Also, $\,x^3-4x+3 \\=x^3-x^2+x^2-x-3x+3\\=x^2(x-1)+x(x-1)-3(x-1)\\=(x-1)(x^2+x-3)\rightarrow(2)$

 $\,\lim_{x \to 1}\,\frac{x^2-3x+2}{x^3-4x+3}\\=\lim_{x \to 1}\frac{(x-2)(x-1)}{(x-1)(x^2+x-3)}\\=\lim_{x \to 1}\frac{x-2}{x^2+x-3}~~~[\text{By (1),(2)}]\\=\frac{1-2}{1^2+1-3}\\=1\,\,\text{(ans.)}$

$\,5.\,$ Evaluate : $\,\lim_{x \to -2}\,\frac{x^3-7x-6}{x^4+5x-6}$

Sol. At first, we factorize $\,x^3-7x-6\,$ and $\,x^4+5x-6$.

$\,~~~~x^3-7x-6\\=x^3+1-7x-7\\=(x+1)(x^2-x+1)-7(x+1)\\=(x+1)(x^2-x+1-7)\\=(x+1)(x^2-x-6)\\=(x+1)(x^2-3x+2x-6)\\=(x+1)\{x(x-3)+2(x-3)\}\\=(x+1)(x-3)(x+2)\rightarrow(1)$

Again, $\,x^4+5x-6 \\=x^4-1+5x-5\\=(x^2+1)(x^2-1)+5(x-1)\\=(x^2+1)(x+1)(x-1)+5(x-1)\\=(x-1)[(x^2+1)(x+1)+5]\rightarrow(2)\\ \text{Now},\, (x^2+1)(x+1)+5\\=x^3+x^2+x+6\\=x^3+2x^2-x^2-2x+3x+6\\=x^2(x+2)-x(x+2)+3(x+2)\\=(x+2)(x^2-x+3)\rightarrow(3) $

Hence, by $\,(2)\,$ and $\,(3),\,$ we get,

$x^4+5x-6 \\=(x-1)(x+2)(x^2-x+3)\rightarrow(4)$

So, $\lim_{x \to -2}\frac{x^3-7x-6}{x^4+5x-6}\\=\lim_{x \to -2}\frac{(x+1)(x-3)(x+2)}{(x-1)(x+2)(x^2-x+3)}~~[\text{By (1), (4)}]\\=\lim_{x \to -2}\frac{(x+1)(x-3)}{(x-1)(x^2-x+3)}\\=\frac{(-2+1)(-2-3)}{(-2-1)(4+2+3)}\\=\frac{(-1)(-5)}{(-3)\times 9}\\=-\frac{5}{27}\,\,\text{(ans.)}$