Question

(i) If $f\left(x\right)=\{\begin{array}{cc}\frac{x-\left|x\right|}{x}& ifx\ne 0\\ 2& ifx=0\end{array}$, show that $\underset{x\to 0}{lim}f\left(x\right)$ does not exist.

(ii) Evaluate $\underset{x\to 0}{lim}\frac{sinx-2xin3x+sin5x}{x}.$

Or

(i) Find the derivative of $\frac{(x-1)(x-2)}{(x-3)(x-4)}.$

(ii) Differentiate $x{e}^x$ by using first principle.

Answer 1

(i) Given, $f\left(x\right)=\{\begin{array}{cc}\frac{x-\left|x\right|}{x}& ifx\ne 0\\ 2& ifx=0\end{array}$

Now, limite of f(x) at x = 0,

$LHL=\underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}\frac{x-\left|x\right|}{x}$

= $\underset{h\to 0}{lim}\frac{(0-h)-|0-h|}{0-h}$

$[putx=0-hasx\to 0,thenh\to 0]$

= $\underset{x\to 0}{lim}\frac{-h-h}{-h}=\underset{x\to 0}{lim}\frac{2h}{h}=2$

$RHL=\underset{x\to 0}{lim},f\left(x\right)$

= $\underset{x\to 0}{lim},\frac{x-\left|x\right|}{x}$

= $\underset{h\to 0}{lim}\frac{(0+h)-|0+h|}{0+h}$

$[putx=0+hasx\to 0,thenh\to 0]$

= $\underset{h\to 0}{lim}\frac{h-h}{h}=0$

$\because LHL\ne RHL$

Hence, $\underset{x\to 0}{lim}f\left(x\right)$ does not exist.

(ii) We have, $\underset{x\to 0}{lim}\frac{sinx-2sin3x+sin5x}{x}$

= $\underset{x\to 0}{lim}\frac{(sinx+sin5x)-2sin3x}{x}$

= $\underset{x\to 0}{lim}\frac{2sin\left(\frac{x+5x}{2}\right)cos\left(\frac{x-5x}{2}\right)-2sin3x}{x}$

$[\because sinA+sinB=2sin\left(\frac{A+B}{2}\right)cos\left(\frac{A-B}{2}\right)]$

= $\underset{x\to 0}{lim}\frac{2sin3xcos2x-2sin3x}{x}$

= $\underset{x\to 0}{lim}\frac{2sin3x(cos2x-1)}{x}$

= $\underset{x\to 0}{lim}\frac{2sin3x}{3x}\times 3\times \underset{x\to 0}{lim}(cos2x-1)$

= $2\times 3\times (1-1)$ $[\because \underset{h\to 0}{lim}\frac{sinh}{h}=1]$

= $6\times 0=0$

Or

(i) Let $y=\frac{(x-1)(x-2)}{(x-3)(x-4)}$

= $\frac{x^2-3x+2}{x^2-7x+12}$

On differentiating both sides w.r.t.x, we get

$\frac{dy}{dx}=\frac{\left[\right(x^2-7x+12\left)\frac{d}{dx}\right(x^2-3x+2)-(x^2-3x+2\left)\frac{d}{dx}\right(x^2-7x+12\left)\right]}{(x^2-7x+12{)}^2}$

[using quotient rule of derivative]

$\therefore \frac{dy}{dx}=\frac{\left[\right(x^2-7x+12\left)\right(2x-3)-(x^2-3x+2\left)\right(2x-7\left)\right]}{(x^2-7x+12{)}^2}$

= $\frac{\left[2x\right(x^2-7x+12)-3(x^2-7x+12)-2x(x^2-3x+2)+7(x^2-3x+2\left)\right]}{(x^2-7x+12{)}^2}$

= $\frac{[2x^3-14x^2+24x-3x^2+21x-36-2x^3+6x^2-4x+7x^2-21x+14]}{(x^2-7x+12{)}^2}$

= $\frac{[-14x^2-3x^2+6x^2+7x^2+24x+21x-4x-21x-36+14]}{(x^2-7x+12{)}^2}$

= $\frac{-4x^2+20x-22}{(x^2-7x+12{)}^2}$

(ii) Let $f\left(x\right)=x{e}^x.$

Then, $f(x+h)=(x+h)e^{x+h}$

$\therefore \frac{d}{dx}\left[f\right(x\left)\right]=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$

= $\underset{h\to 0}{lim}\frac{(x+h)e^{x+h}-x{e}^x}{h}$

= $\underset{h\to 0}{lim}\frac{(x^{x+h}-x{e}^x)+h{e}^{x+h}}{h}$

= $\underset{h\to 0}{lim}\left[\frac{xe{}^x(e^h-1)}{h}\right]+\underset{h\to 0}{lim}{e}^{x+h}$

= $x{e}^x\underset{h\to 0}{lim}\left(\frac{e^h-1}{h}\right)+\underset{h\to 0}{lim}{e}^{x+h}$

= $x{e}^x\times 1+e^{x+0}$ $[\because \underset{h\to 0}{lim}\frac{e^h-1}{h}=1]$

= $x{e}^x{+}^{.}{e}^x=e^x(x+1)$