Question

$\underset{x\to 7/2}{lim}(2x^2-9x+8{)}^{\cot (2x-7)}$ is equal to

• A. $e^{\frac{5}{2}}$
• B. $e^{-\frac{5}{2}}$
• C. $e^{\frac{7}{2}}$
• D. $e^{\frac{3}{2}}$

Answer 1

The correct option is A $e^{\frac{5}{2}}$

Here,
$\underset{x\to 7/2}{lim}(2x^2-9x+8{)}^{\cot (2x-7)}[1^{\infty }\text{form}]$
$=e^{\underset{x\to 7/2}{lim}\left(\right(2x^2-9x+8)-1)\cdot \cot (2x-7)}$
$=e^{\underset{x\to 7/2}{lim}(2x^2-9x+7)\cdot \cot (2x-7)}$
$=e^{\underset{x\to 7/2}{lim}\frac{(2x-7)(x-1)}{\tan (2x-7)}}$
$=e^{\underset{x\to 7/2}{lim}(x-1)}(\because \underset{x\to 7/2}{lim}\frac{\tan (2x-7)}{(2x-7)}=1)$
$=e^{5/2}$