Question

evaluate the following limits :

$\sf{ lim_{x \to1}} \frac{(2x - 3)(\sqrt{x} - 1)}{ {3x}^{2} + 3x - 6}$
​

Answer 1

Answer:

This implies that

x2+2ax=4x−4a−13

or

x2+2ax−4x+4a+13=0

or

x2+(2a−4)x+(4a+13)=0

Since the equation has just one solution instead of the usual two distinct solutions, then the two solutions must be same i.e. discriminant = 0.

Hence we get that

(2a−4)2=4⋅1⋅(4a+13)

or

4a2−16a+16=16a+52

or

4a2−32a−36=0

or

a2−8a−9=0

or

(a−9)(a+1)=0

So the values of a are −1 and 9.

Answer 2

Answer:

$\sf{ lim_{x \to1}} \frac{(2x - 3)(\sqrt{x} - 1)}{ {3x}^{2} + 3x - 6} \\ \\ \sf \color{red}{ factorize \: the \: denominator \: by \: middle \: term \: splitting}\\ \\ : \implies \sf{lim_{x \to \: 1} \frac{(2x - 3)( \sqrt{x} - 1) }{3 {x}^{2} - 3x + 6x - 6} } \\ \\ \sf{ : \implies lim_{x \to \: 1} = \frac{(2x - 3)( \sqrt{x} - 1) }{3x(x - 1) + 6(x - 1)} }</p><p> \\ \\ \sf{ : \implies lim_{x \to \: 1} \frac{(2x - 3)( \sqrt{x} - 1)}{(3x + 6) (x - 1)} } \\ \\ \sf{ : \to \color{green}\:( x - 1) = (\sqrt{x} - 1)( \sqrt{x} + 1)} \\ \\ \sf{ : \implies \: lim_{x \to \: 1} \frac{(2x - 3)( \sqrt{x} - 1)}{(3x + 6)( \sqrt{x} - 1)( \sqrt{x} + 1)} } \\ \\ \sf{ : \implies lim_{x \to \: 1} \frac{(2x - 3)( \sqrt{x} - 1)}{(3x + 6)( \sqrt{x} + 1)} } \\ \\ \sf \color{gold}{ \to \: subsitute \: x \: as \: 1} \\ \\ : \implies \frac{2 \times 1 - 3}{(3 \times 1 + 6)(1 + 1)} \\ \\ \sf{ : \implies \: \color{blue}\frac{ - 1}{18} }$