Question

evaluate the limit...​

Answer 1

Answer:

Answer is  -4 / 3

Step-by-step explanation:

We Have :-

$\lim_{x \to 2} (\frac{3^{x}+3^{3-x}-12}{3^{3-x}-3^{\frac{x}{2} }} )$

Solution :-

There are 2 ways to do this :-

1 Way :-

$\lim_{x \to 2} (\frac{3^{x}+3^{3-x}-12}{3^{3-x}-3^{\frac{x}{2} }} )\\\\ \lim_{t \to 9} ( \frac{t + \frac{27}{t}- 12 }{\frac{27}{t}-t^{\frac{3}{2} } } )\\\\ \lim_{t \to 9} ( \frac{t^{2}+27-12t}{27-t^{\frac{3}{2} }} )\\\\Applying\ l'Hopital\\\\ \lim_{t \to 9} ( \frac{2t -12}{\frac{-3}{2}\sqrt{t} } )$

$Applying\ Value\\\\(\frac{18-12}{\frac{-3}{2}*3 } )\\\\\frac{12}{-9}\\\\ \frac{-4}{3}$

2 Way :-

$\lim_{x \to 2} (\frac{3^{x}+3^{3-x}-12}{3^{3-x}-3^{\frac{x}{2} }} )\\\\Applying\ l'Hopital\\\\ \lim_{x \to 2 } ( \frac{3^{x} log 3+3^{3-x}log 3(-1)}{-3^{3-x}log3-\frac{-3^{\frac{x}{2} }}{2} log 3})$

$\lim_{x \to 2} ( \frac{3^{x}-3^{3-x}}{-3^{3-x}-\frac{3^{\frac{x}{2}} }{2} } )\\\\Applying\ Value\\\\= \frac{9-3}{-3-\frac{3}{2} } \\\\= \frac{6}{-9}*2 \\\\= -\frac{4}{3}$

Answer is  -4 / 3

Answer 2

Answer:

use quadratic , factorise a^3-b^3 , a^2-b^2 , without L hoptal rule