Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

\lim_{x\rightarrow 0}\left ( \frac{3x^{2}+2}{7x^{2}+2} \right )^{1/x^{2}}is equal to :
 
Option: 1 e
Option: 2 \frac{1}{e^{2}}
Option: 3 \frac{1}{e}
Option: 4 e^{2}
 

Answers (1)

best_answer

 

 

Limits of the form (1 power infinity) -

Limits of the form 1∞  (1 power infinity)

To find the limit of the exponential of form 1∞ , we will use the following results  

If \lim_{x\rightarrow a}\;f(x)=\lim_{x\rightarrow a}\;g(x)=0

Then,

\lim_{x\rightarrow a}\;\left [1+f(x) \right ]^{\frac{1}{g(x)}}=e^{\lim_{\;x\rightarrow a}\frac{f(x)}{g(x)}} .

Or

When,  \lim_{x\rightarrow a}\;f(x)=1 and \lim_{x\rightarrow a}\;g(x)=\infty

Then, 

\\\lim_{x\rightarrow a}\;\left [f(x) \right ]^{{g(x)}}=\lim_{x\rightarrow a}\;\left [1+f(x)-1 \right ]^{{g(x)}}\\\\\text{}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=e^{\lim_{\;x\rightarrow a}{\left (f(x)-1 \right )}{g(x)}}

Some particular cases  

\\\text{(a)}\;\;\;\lim_{x\rightarrow 0}\;(1+x)^{\frac{1}{x}}=e\\\\\text{(b)}\;\;\;\lim_{x\rightarrow 0}\;\left (1+\frac{1}{x} \right )^{x}=e\\\\\text{(c)}\;\;\;\lim_{x\rightarrow 0}\;(1+cx)^{\frac{1}{x}}=e^c\\\\\text{(d)}\;\;\;\lim_{x\rightarrow 0}\;\left (1+\frac{c}{x} \right )^{x}=e^c

-

 

  

 

 

Limits of the form (1 power infinity) -

Limits of the form 1∞  (1 power infinity)

To find the limit of the exponential of form 1∞ , we will use the following results  

If \lim_{x\rightarrow a}\;f(x)=\lim_{x\rightarrow a}\;g(x)=0

Then,

\lim_{x\rightarrow a}\;\left [1+f(x) \right ]^{\frac{1}{g(x)}}=e^{\lim_{\;x\rightarrow a}\frac{f(x)}{g(x)}} .

Or

When,  \lim_{x\rightarrow a}\;f(x)=1 and \lim_{x\rightarrow a}\;g(x)=\infty

Then, 

\\\lim_{x\rightarrow a}\;\left [f(x) \right ]^{{g(x)}}=\lim_{x\rightarrow a}\;\left [1+f(x)-1 \right ]^{{g(x)}}\\\\\text{}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=e^{\lim_{\;x\rightarrow a}{\left (f(x)-1 \right )}{g(x)}}

Some particular cases  

\\\text{(a)}\;\;\;\lim_{x\rightarrow 0}\;(1+x)^{\frac{1}{x}}=e\\\\\text{(b)}\;\;\;\lim_{x\rightarrow 0}\;\left (1+\frac{1}{x} \right )^{x}=e\\\\\text{(c)}\;\;\;\lim_{x\rightarrow 0}\;(1+cx)^{\frac{1}{x}}=e^c\\\\\text{(d)}\;\;\;\lim_{x\rightarrow 0}\;\left (1+\frac{c}{x} \right )^{x}=e^c

-

 

 

\\\lim _{x\to 0}\left(\frac{3x^2+2}{\:7x^2+2}\right)^{\frac{1}{x^2}}=\lim _{x\to 0}\left(1+\frac{3x^2+2}{\:7x^2+2}-1\right)^{\frac{1}{x^2}}\\\lim_{x\rightarrow 0}\left ( 1+\frac{-4x^2}{7x^2+2} \right )^{1/x^2}=e^{\lim _{x\to 0}\left(-\frac{4x^2}{7x^2+2}\right)\left(\frac{1}{x^2}\right)}\\=\frac{1}{e^2}

 

Correct Option (2)

Posted by

Kuldeep Maurya

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

JEE Main 2024: NTA debars 39 candidates for 3 years on account of using unfair means
anam-khan

JEE Main 2024 session 2 paper 2 answer key objection window closes today; fees
anam-khan

JEE Main 2024 Topper: Farmer’s son Gajare Nilkrishna says ‘keep questioning until you understand’

Latest Question