Question

How do you use the formal definition of limit to prove $\lim \left( \dfrac{x}{x-3} \right)=1$ x approaches to infinity?

Answer 1

Hint: In this question, we have to prove the limit function. Thus, we will use the formal definition of the limit. As we know, the formal definition of limit is; $\displaystyle \lim_{x \to \infty }f\left( x \right)=L$ if and only if for every $\varepsilon >0$ , there is a M such that for all x, if $x>M$ , then $|f\left( x \right)-L|<\varepsilon $ . Thus, in this problem, we will let $M=\dfrac{3}{\varepsilon }+3$ , such that $M>3$ . Then, we will substitute the value of M in the given inequation $x>M$, to get the value of epsilon. Then, we will solve $\left| \dfrac{x}{x-3}-1 \right|$ to get the value be less than epsilon, to get the required solution for the problem.

Complete step-by-step solution:
According to the problem, we have to prove the limit function.
Thus, we will use the formal definition of the limit.
Let us first suppose that there is a M, such that $M=\dfrac{3}{\varepsilon }+3$. Now, we will solve this equation to get the value of epsilon, so we will first subtract 3 on both sides in the above equation. Also, we know that $x>M$, thus we get
$\Rightarrow x-3>\dfrac{3}{\varepsilon }+3-3$
As we know, the same terms with opposite signs cancel out each other, thus we get
$\Rightarrow x-3>\dfrac{3}{\varepsilon }$
Now, we will divide 3 on both sides in the above equation, we get
$\Rightarrow \dfrac{x-3}{3}>\dfrac{3}{3\varepsilon }$
On furthers simplification, we get
$\Rightarrow \dfrac{x-3}{3}>\dfrac{1}{\varepsilon }$
Now, we will take the reciprocal of the above equation, we get
$\Rightarrow \varepsilon >\dfrac{3}{x-3}$ --------- (1)
Since, the function given to us is $f\left( x \right)=\dfrac{x}{x-3}$ , so we will now take the modulus and subtract the same function by 1, we get
$\Rightarrow \left| \dfrac{x}{x-3}-1 \right|$
Now, we will take LCM of the denominator in the above function, we get
$\Rightarrow \left| \dfrac{x-\left( x-3 \right)}{x-3} \right|$
Now, we will open the brackets of the above equation, we get
$\Rightarrow \left| \dfrac{x-x+3}{x-3} \right|$
As we know, the same terms with opposite signs cancel out each other, thus we get
$\Rightarrow \left| \dfrac{+3}{x-3} \right|$
Also, from equation (1), we get that
$\Rightarrow \left| \dfrac{+3}{x-3} \right|<\varepsilon $
Therefore, the definition of limit says that $\displaystyle \lim_{x \to \infty }f\left( x \right)=L$ if and only if for every $\varepsilon >0$ , there is a M such that for all x, if $x>M$ , then $|f\left( x \right)-L|<\varepsilon $ . Thus, we get
$\displaystyle \lim_{x \to \infty }\left( \dfrac{x}{x-3} \right)=1$
Hence proved.

Note: While solving this problem, do mention every step properly to avoid mistakes and confusion. Also, mention the definition of limit before starting your solution.