Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Find the value of following limit:  <img alt="\lim _{x \rightarrow \infty} \frac{-23 x^4+45 x^3+8 x^2}{7 x^3+18 x^2+5}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Clim%20_%7Bx%20%5

Find the value of following limit:  \lim _{x \rightarrow \infty} \frac{-23 x^4+45 x^3+8 x^2}{7 x^3+18 x^2+5}

 

Option: 1

0


Option: 2

-\infty


Option: 3

\infty


Option: 4

Limit does not exist.


Answers (1)

best_answer

\lim _{x \rightarrow \infty} \frac{-23 x^4+45 x^3+8 x^2}{7 x^3+18 x^2+5}

For solving this equation, a common result can be used:
If  a, b are positive integers and m_0, n_0 \neq 0 are non-zero real numbers, then

\lim _{x \rightarrow \infty} \frac{m_0 x^a+m_1 x^{a-1}+\ldots+m_{a-1} x+m_a}{n_0 x^b+n_1 x^{b-1}+\ldots+n_{b-1} x+n_b}= \begin{cases}0 & \text { if } a<b \\ \frac{m_0}{n_0} & \text { if } a=b \\ \infty & \text { if } a>b \text { when } m_0 n_0>0 \\ -\infty & \text { if } a>b \text { when } m_0 n_0<0\end{cases}

Now, \mathrm{a}=4 and \mathrm{b}=3 , which means \mathrm{a}>\mathrm{b} andm_0 n_0<0, hence according to result:

\lim _{x \rightarrow \infty} \frac{-23 x^4+45 x^3+8 x^2}{7 x^3+18 x^2+5}=-\infty

Posted by

avinash.dongre

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

BTech aspirant from north-east or UT? Apply for CSAB NEUT counselling 2025 today
anam-khan

JoSAA round 2 cut-off 2025 out; BTech closing ranks for top engineering courses at IITs
anam-khan

Goa Engineering Admission 2025: 1,415 eligible for BE seats; merit list on June 27

Latest Question