Get Answers to all your Questions

header-bg qa

Evaluate the limit  \lim _{x \rightarrow \infty} \frac{f(3 x)}{f(2 x)}, for f:R\rightarrow R is an increasing function and  \lim _{x \rightarrow \infty} \frac{f(5 x)}{f(2 x)}=1

Option: 1

\pm \pi


Option: 2

1


Option: 3

0


Option: 4

\pi


Answers (1)

best_answer

The Sandwich theorem also known as squeeze play theorem applicable for any three real valued functions f(x),g(x) and h(x)  that shares the same domain such that f(x)\leq g(x)\leq h(x) for \forall x  in the domain of definition states the following:For some real value of a, if \lim _{x \rightarrow a} f(x)=l=\lim _{x \rightarrow a} h(x) then \lim _{x \rightarrow a} f(x)=l

Note that the provide function F:R\rightarrow R is an increasing function.The stated limit is 

\lim _{x \rightarrow \infty} \frac{f(5 x)}{f(2 x)}=1 ____________(i)

Observe the following for F:R\rightarrow R  is an increasing function.

As (2 x)<(3 x)<(4 x)<(5 x),, it is evident that 

f(2 x)<f(3 x)<f(4 x)<f(5 x)

So, the following inequality is of importance

f(2 x)<f(3 x)<f(5 x) -------------(ii)

Now, divide the inequality (ii) by the function f(2x) that still preserves the inequality.

\begin{aligned} & \frac{f(2 x)}{f(2 x)}<\frac{f(3 x)}{f(2 x)}<\frac{f(5 x)}{f(2 x)} \\ & 1<\frac{f(3 x)}{f(2 x)}<\frac{f(5 x)}{f(2 x)} \end{aligned}--------------(iii)

Apply the squeeze theorem to the inequality (iii) and use the equation (i).

\begin{aligned} & \lim _{x \rightarrow \infty} 1 \leq \lim _{x \rightarrow \infty} \frac{f(3 x)}{f(2 x)} \leq \lim _{x \rightarrow \infty} \frac{f(5 x)}{f(2 x)} \\ & 1 \leq \lim _{x \rightarrow \infty} \frac{f(3 x)}{f(2 x)} \leq 1 \end{aligned}

Therefore, the required value of the limit is

\lim _{x \rightarrow \infty} \frac{f(3 x)}{f(2 x)}=1

Posted by

Rishabh

View full answer

JEE Main high-scoring chapters and topics

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