logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XI  >>  Math  >>  Limits and Derivatives
0 votes

Evaluate the following limits \[\] $ \lim\limits_{x \to 3} \large\frac{x^4-81}{2x^2-5x-3}$

$\begin{array}{1 1}\large\frac{108}{7} \\ 0 \\ \infty \\ \large\frac{18}{7} \end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
At $ x = 2$ the rational function takes the form $ \large\frac{0}{0}$
$ \Rightarrow \lim\limits_{x \to 3} \large\frac{x^4-81}{2x^2-5x-3}$ $ = \lim\limits_{x \to 3} \large\frac{(x-3)(x+3)(x^2+9)}{(x-3)(2x+1)}$
$\qquad = \lim\limits_{x \to 3} \large\frac{(x+3)(x^2+9)}{2x+1}$
$ \qquad = \large\frac{(3+3)(3^2+9)}{2(3)+1}$
$\qquad = \large\frac{6 \times 18}{7}$ $ = \large\frac{108}{7}$
answered Apr 3, 2014 by thanvigandhi_1
edited May 16, 2014 by balaji.thirumalai
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...