3.1 Basic Limits - Introduction to Calculus

Basic Limits

Suppose we have a function $y=f(x)$ The limit of $f(x)$ as $x$ approaches the number $a$ describes the value that $f(x)$ approaches, and is denoted $\lim\limits_{x\to a}f(x)$

For example, suppose we wanted to determine $\lim\limits_{x\to3}x^2$

$x$ approaches 3 from the left	$x$ approaches 3 from the right
$f(2.9)=8.41$	$f(3.1)=9.61$
$f(2.99)=8.9401$	$f(3.01)=9.0601$
$f(2.999)=8.994001$	$f(3.001)=9.006001$

No matter which way we approach from, as $x$ gets close to 3, we can see that $x^2$ gets very close to 9.

Follow the original example, then change the target. The two inputs stay different from the target even as their outputs approach the same value.

\displaystyle f(x)=x^2

Target a3Closeness · larger means nearer1

\displaystyle h=10^{-1}=0.1

Approach	$\displaystyle x$	$\displaystyle f(x)$
From the left	2.9	8.41000000
From the right	3.1	9.61000000

\displaystyle \lim_{x\to 3}x^2=9

QUESTION

\displaystyle \lim_{x\to0}f(x)=1

but

\displaystyle f(0)=-1

, what is the limit?

Find $\displaystyle \lim\limits_{x\to0}f(x)$ where

\displaystyle f(x)=\begin{cases}\cos(x)&x\ne0\\[.4em]-1&x=0\end{cases}

\displaystyle f(x)=\cos x,\quad x\ne0

Law of Limits

THEOREM

Law of Limits

Suppose both $\lim\limits_{x\to a}f(x)$ and $\lim\limits_{x\to a}g(x)$ exist as finite real numbers. Then

\begin{aligned} \lim_{x\to a}(f+g)(x)&=\lim_{x\to a}f(x)+\lim_{x\to a}g(x)\\[.65em] \lim_{x\to a}(f-g)(x)&=\lim_{x\to a}f(x)-\lim_{x\to a}g(x)\\[.65em] \lim_{x\to a}(fg)(x)&=\lim_{x\to a}f(x)\cdot\lim_{x\to a}g(x)\\[.65em] \lim_{x\to a}\left(\frac fg\right)(x)&=\frac{\lim\limits_{x\to a}f(x)}{\lim\limits_{x\to a}g(x)} \end{aligned}

The quotient rule additionally requires $\lim\limits_{x\to a}g(x)\ne0$

The laws apply only when their assumptions hold. For an oscillating function, a different idea may help: if a function lies between two bounds that approach the same value, it must approach that value too. This is the squeeze theorem.

Evaluate $\displaystyle \lim\limits_{x\to0}x\sin\left(\frac1x\right)$

One-sided Limits

$x$ can approach the value $a$ in two ways:

Left-hand limit: $\lim\limits_{x\to a^-}f(x)$
Right-hand limit: $\lim\limits_{x\to a^+}f(x)$

We can tighten our definition of the limit of a function $f(x)$ as $x$ approaches $a$ If the left-hand and right-hand limits of $f(x)$ are both equal to a number $L$ , then we say that $\lim\limits_{x\to a}f(x)$ exists and is equal to $L$

The initial settings are the practice function. Change the height of the right branch to make the one-sided limits agree; then decide whether the isolated function value matters.

\displaystyle f(x)=x,\quad x<1

\displaystyle f(x)=\sqrt{x-1}+(0),\quad x>1

Right-branch vertical shift0Isolated value f(1)-1Closeness · larger means nearer1

\displaystyle h=10^{-1}=0.1

Approach	$\displaystyle x$	$\displaystyle f(x)$
From the left	0.9	0.90000000
From the right	1.1	0.31622777

\displaystyle \lim_{x\to1^-}f(x)=1,\qquad\lim_{x\to1^+}f(x)=0

The two-sided limit does not exist: the one-sided limits differ.

\displaystyle f(1)=-1

QUESTION

Suppose

\displaystyle \lim_{x\to a^-}f(x)=2

and

\displaystyle \lim_{x\to a^+}f(x)=3

What can you conclude?

Consider the piecewise function given by

\displaystyle f(x)=\begin{cases}x&x<1\\[.4em]-1&x=1\\[.4em]\sqrt{x-1}&x>1\end{cases}

Find the following values and decide whether the two-sided limit exists.

(a) $\displaystyle \lim\limits_{x\to1^-}f(x)$

(b) $\displaystyle \lim\limits_{x\to1^+}f(x)$

(d) Does $\displaystyle \lim\limits_{x\to1}f(x)$ exist?

Basic Limits

Suppose we have a function $y=f(x)$ The limit of $f(x)$ as $x$ approaches the number $a$ describes the value that $f(x)$ approaches, and is denoted $\lim\limits_{x\to a}f(x)$

For example, suppose we wanted to determine $\lim\limits_{x\to3}x^2$

$x$ approaches 3 from the left	$x$ approaches 3 from the right
$f(2.9)=8.41$	$f(3.1)=9.61$
$f(2.99)=8.9401$	$f(3.01)=9.0601$
$f(2.999)=8.994001$	$f(3.001)=9.006001$

No matter which way we approach from, as $x$ gets close to 3, we can see that $x^2$ gets very close to 9.

Follow the original example, then change the target. The two inputs stay different from the target even as their outputs approach the same value.

\displaystyle f(x)=x^2

Target a3Closeness · larger means nearer1

\displaystyle h=10^{-1}=0.1

Approach	$\displaystyle x$	$\displaystyle f(x)$
From the left	2.9	8.41000000
From the right	3.1	9.61000000

\displaystyle \lim_{x\to 3}x^2=9

QUESTION

\displaystyle \lim_{x\to0}f(x)=1

but

\displaystyle f(0)=-1

, what is the limit?

Find $\displaystyle \lim\limits_{x\to0}f(x)$ where

\displaystyle f(x)=\begin{cases}\cos(x)&x\ne0\\[.4em]-1&x=0\end{cases}

\displaystyle f(x)=\cos x,\quad x\ne0

Law of Limits

THEOREM

Law of Limits

Suppose both $\lim\limits_{x\to a}f(x)$ and $\lim\limits_{x\to a}g(x)$ exist as finite real numbers. Then

\begin{aligned} \lim_{x\to a}(f+g)(x)&=\lim_{x\to a}f(x)+\lim_{x\to a}g(x)\\[.65em] \lim_{x\to a}(f-g)(x)&=\lim_{x\to a}f(x)-\lim_{x\to a}g(x)\\[.65em] \lim_{x\to a}(fg)(x)&=\lim_{x\to a}f(x)\cdot\lim_{x\to a}g(x)\\[.65em] \lim_{x\to a}\left(\frac fg\right)(x)&=\frac{\lim\limits_{x\to a}f(x)}{\lim\limits_{x\to a}g(x)} \end{aligned}

The quotient rule additionally requires $\lim\limits_{x\to a}g(x)\ne0$

Evaluate $\displaystyle \lim\limits_{x\to0}x\sin\left(\frac1x\right)$

One-sided Limits

$x$ can approach the value $a$ in two ways:

Left-hand limit: $\lim\limits_{x\to a^-}f(x)$
Right-hand limit: $\lim\limits_{x\to a^+}f(x)$

The initial settings are the practice function. Change the height of the right branch to make the one-sided limits agree; then decide whether the isolated function value matters.

\displaystyle f(x)=x,\quad x<1

\displaystyle f(x)=\sqrt{x-1}+(0),\quad x>1

Right-branch vertical shift0Isolated value f(1)-1Closeness · larger means nearer1

\displaystyle h=10^{-1}=0.1

Approach	$\displaystyle x$	$\displaystyle f(x)$
From the left	0.9	0.90000000
From the right	1.1	0.31622777

\displaystyle \lim_{x\to1^-}f(x)=1,\qquad\lim_{x\to1^+}f(x)=0

The two-sided limit does not exist: the one-sided limits differ.

\displaystyle f(1)=-1

QUESTION

Suppose

\displaystyle \lim_{x\to a^-}f(x)=2

and

\displaystyle \lim_{x\to a^+}f(x)=3

What can you conclude?

Consider the piecewise function given by

\displaystyle f(x)=\begin{cases}x&x<1\\[.4em]-1&x=1\\[.4em]\sqrt{x-1}&x>1\end{cases}

Find the following values and decide whether the two-sided limit exists.

(a) $\displaystyle \lim\limits_{x\to1^-}f(x)$

(b) $\displaystyle \lim\limits_{x\to1^+}f(x)$

(d) Does $\displaystyle \lim\limits_{x\to1}f(x)$ exist?

Basic Limits

Basic Limits

Approach from both sides

KNOWLEDGE QUIZ

A limit with a different function value

Law of Limits

Law of Limits

Squeeze an oscillating function

One-sided Limits

Compare the left and right limits

KNOWLEDGE QUIZ

Compare one-sided limits

Basic Limits

Basic Limits

Approach from both sides

KNOWLEDGE QUIZ

A limit with a different function value

Law of Limits

Law of Limits

Squeeze an oscillating function

One-sided Limits

Compare the left and right limits

KNOWLEDGE QUIZ

Compare one-sided limits