Home Calculus How to find the limit of a function?

How to find the limit of a function?

how-to-find-the-limit-of-a-function

We propose simple techniques to learn you how to find the limit of a function. We first recall what a limit of a function is. Then give some properties of the limit as well as several examples in the form of exercises with detailed solutions.

Generalities on limits of functions.

let $a$ a real number and $f:\mathbb{R}\setminus \{a\}\to\mathbb{R}$ be function.

  • We say that $f$  has a limit $\ell\in \mathbb{R}$ at the point $a$ if for any small open interval $V_\ell$ centered in $\ell$, there exists a small open interval $I_a$ centered in $a$ such that $f(I_a)\subset V_\ell$. That is, for any small real number $\varepsilon>0,$ there exists a small real number $\alpha>0$ such that, for any $x,$ $|x-a|<\alpha$ implies that $|f(x)-\ell|<\varepsilon$. Here we can take $I_a=(a-\alpha,a+\alpha)$ and $V_\ell=(\ell-\varepsilon,\ell+\varepsilon)$. In this case, we write $$\lim_{x\to a}f(x)=\ell.$$
  • The function $f$ has a right limit $\ell$ at $a$, if for any $\varepsilon>0,$ there exists $\alpha>0,$ such that, for any $x$, $a<x<a+\alpha$ implies that $|f(x)-\ell|<\varepsilon$. In this case, we write $$\lim_{x\to a^+}f(x)=\ell.$$
  • We say that $f$ has a left limit $\ell$ at $a$, if for any $\varepsilon>0,$ there exists $\alpha>0,$ such that, for any $x$, $a-\alpha<x<a$ implies that $|f(x)-\ell|<\varepsilon$. In this case, we write $$\lim_{x\to a^-}f(x)=\ell.$$
  • The function $f$ has a limit $+\infty$ at the point $a$ if for any sufficiently large $A>0,$ there exist a small $\alpha>0$ such that, for any $x,$ $|x-a|<\alpha$ implies $f(x)>A$. In this case, we write $$ \lim_{x\to a}f(x)=+\infty.$$
  • We say that $f$ has a limit $-\infty$ at the point $a$ if for any sufficiently large $A>0,$ there exist a small $\alpha>0$ such that, for any $x,$ $|x-a|<\alpha$ implies $f(x)<-A$. In this case, we write $$ \lim_{x\to a}f(x)=-\infty.$$

Proposition: The function $f$ admits a limit $\ell$ at the point $a$ if and only if it admits the right and left limits at $a$ and these limits are equal to $\ell$.

Similar to the limits of the sequence we have the following result.

The squeeze theorem for functions: Assume that there exist three real functions satisfying $h\le f\le g$ and that $$\lim_{x\to a}h(x)=\lim_{x\to a}g(x)=\ell.$$ Then the function has the limit $\ell$ at the point $a$.

Relation with sequences limits: The function has the limit $\ell$ at the point $a$ if and only if for any sequence $(u_n)_n$ such that $u_n$ converges to $a$, the sequence image $f(u_n)$ converges to $\ell$.

This result is very useful if we want to show that a function has no limit at a point. In fact, it suffices to find two sequences $(u_n)_n$ and $(v_n)_n$ which converge to $a,$ while the image sequences $f(u_n)_n$ and $f(v_n)$ converge to different real numbers.

How to find the limit of a function?

Exercise: Determine the limits of the following functions

  1. $ f(x)=\sqrt{x+1}-\sqrt{x}\qquad (x\to+\infty) $
  2. $ g(x)= \displaystyle\frac{x+\cos x}{x+\sin x}\qquad (x\to +\infty) $
  3. $h(x)=\displaystyle \sqrt{x+\sqrt x}-\sqrt{x}\qquad (x\to +\infty) $
  4. $ \varphi(x)= \displaystyle\sin(x)\sin\left(\frac{1}{x}\right)\qquad (x\to +\infty) $

Solution: 1) Observe that $\sqrt{x+1}$ and $\sqrt{x} $ goes to $+\infty$ as $x\to+\infty,$ then we have an indeterminate form. So you have to rewrite $f$ in another form. We recall that if $a,b$ are in $[0,+\infty)$, then $(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})=a-b$. By using this, we obtain, for any $x>0,$ \begin{align*} f(x)&=\sqrt{x+1}-\sqrt{x}\cr &=\frac{(x+1)-x}{\sqrt{x+1}+\sqrt{x}}\cr =&\frac{1}{\sqrt{x+1}+\sqrt{x}}. \end{align*} We now see that \begin{align*} \lim_{x\to +\infty}f(x)=0. \end{align*}

2) It is well know that the functions $x\mapsto \cos(x)$ and $x\mapsto \sin(x)$ have not limits as $x\to\infty$, these functions are periodic and have values in $[-1,1]$. To compute the limit of $g$ we follow the following technic: For any $x>0,$ we have \begin{align*} g(x)&=\displaystyle\frac{x\left(1+\frac{\cos(x)}{x}\right)}{x\left(1+\frac{\sin(x)}{x}\right)} \cr &=\displaystyle\frac{1+\frac{\cos(x)}{x}}{1+\frac{\sin(x)}{x}} \end{align*} On the other hand, we have \begin{align*} \left|\frac{\cos(x)}{x} \right| \le \frac{1}{x},\qquad \left|\frac{\sin(x)}{x} \right| \le \frac{1}{x}. \end{align*} Then \begin{align*} \lim_{x\to +\infty}\frac{\cos(x)}{x}=0,\qquad \lim_{x\to +\infty}\frac{\sin(x)}{x}=0. \end{align*} Thus \begin{align*} \lim_{x\to +\infty}g(x)=\frac{1}{1}=1. \end{align*}

3) As in the first question, for $x>0$ we write \begin{align*} h(x)&=\frac{\sqrt{x}}{ \sqrt{x+\sqrt x}+\sqrt{x}}\cr & =\frac{\sqrt{x}}{\sqrt{x}\left( \sqrt{\sqrt {x}+1}+1\right)} \cr & =\frac{1}{\sqrt{\sqrt {x}+1}+1} \end{align*} Clearly, \begin{align*} \lim_{x\to +\infty}h(x)=0. \end{align*}

4) The idea is to use the following estimates \begin{align*} |\sin(x)|\le 1\quad\text{and}\quad \left|\sin\left(\frac{1}{x}\right)\right|\le \frac{1}{x}. \end{align*} Then \begin{align*} |\varphi(x)|\le \frac{1}{x}\underset{x\to+\infty}{\longrightarrow} 0. \end{align*} Hence \begin{align*} \lim_{x\to +\infty}\varphi(x)=0. \end{align*}

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