18.01: Single Variable Calculus
Section 2
Limits, Continuity, and Trigonometric Limits
Massachusetts Institute of Technology
Last Edit Date: 06/04/2024
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More about the "rate of change" interpretation of the derivative¶
$$\frac{\Delta y}{\Delta x} \rightarrow \frac{dy}{dx}~\text{as}~\Delta x \rightarrow 0$$
$$\text{Average rate of change}~\rightarrow~\text{Instantaneous rate of change}$$
Examples¶
$q = \text{charge}$, $\frac{dq}{dt} = \text{electrical current}$
$s = \text{distance}$, $\frac{ds}{dt} = \text{speed}$
$T = \text{temperature}$, $\frac{dT}{dx} = \text{temperature gradient}$
Sensitivity of measurements: In GPS, radio signals give us $h$ up to a certain measurement error. The question is how accurately can we measure $L$. To decide, we find $\frac{\Delta L}{\Delta h}$. In the other words, these variables are related to each other. We want to find how a change in one variable affects the other variables.
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Limits and Continuity¶
Easy Limits¶
$$\lim_{x \rightarrow 3} \frac{x^2 + x}{x + 1} = \frac{3^2 + 3}{3 + 1} = \frac{12}{4} = 3$$
With an easy limit, you can get a meaningful answer just by plugging in the limiting value.
Remember,
$$\lim_{x \rightarrow x_0} \frac{\Delta f}{\Delta x} = \lim_{x \rightarrow x_0} \frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x}$$
is never an easy limit, because the denominator $\Delta x = 0$ is not allowed. (The limit $x \rightarrow x_0$ is computed under the implicit assumption that $x \neq x_0$.)
Continuity¶
We say $f(x)$ is continuout at $x_0$ when
$$\lim_{x \rightarrow x_0} f(x) = f(x_0)$$
$$ f(x) = \begin{cases} x + 1 & \text{if } x > 0 \\ -x & \text{if } x \le 0 \end{cases} $$
This discontinuous function is seen in the figure above. For $x > 0$,
$$\lim_{x \rightarrow 0} f(x) = 1$$
but $f(0) = 0$. (One can also say, $f$ is continuous form the left at 0, not the right.)
Removeable Discontinuity¶
Defifinition of removable discontinuity
Right-hand limit: $\lim_{x \rightarrow x_0^{+}} f(x)$ means $\lim_{x \rightarrow x_0} f(x)$ for $x > x_0$.
Left-hand limit: $\lim_{x \rightarrow x_0^{-}} f(x)$ means $\lim_{x \rightarrow x_0} f(x)$ for $x < x_0$.
If $\lim_{x \rightarrow x_0^+} f(x) = \lim_{x \rightarrow x_0^-} f(x)$ but this is not $f(x_0)$, or if $f(x_0)$ is undefined, we say the discontinuity is removable.
For example, $\frac{\sin(x)}{x}$ is defined for $x \neq 0$. We will see that how to evalulate the limit as $x \rightarrow 0$.
Jump Discontinuity¶
$\lim_{x \rightarrow x_0^+}$ for $(x < x_0)$ exists, and $\lim_{x \rightarrow x_0^-}$ for $(x > x_0)$ also exists, but they are NOT equal.
Infinite Discontinuity¶
Right-hand limit: $\lim_{x \rightarrow 0^+} \frac{1}{x} = \infty$
Left-hand limit: $\lim_{x \rightarrow 0^-} \frac{1}{x} = -\infty$
Other (ugly) Discontinuities¶
This funciton does not even go to $\pm \infty$ - it does not make sense to say it goes to anything. For something like this, we say the limit does not exist.
Picturing the Derivative¶
Note that the graph of $f(x)$ does NOT look like the graph of $f'(x)$! (You might also notice that $f(x)$ is an odd function, while $f'(x)$ is an even function. The derivative of an odd function is always even, and vice versa.)
Pumpkin Drop, Part II¶
This time, someone throws a pumpkin over the tallest building on campus.
$$y(t) = 400 - 16t^2,~-5 \le t \le 5$$
The bottom figure shows the derivative ($y'(t)$) of $y(t) = 400 - 16t^2$.
Two Trig Limits¶
In the expressions below, $\theta$ is in radians - NOT degrees.
$$\boxed{\lim_{\theta \rightarrow 0}{\frac{\sin \theta}{\theta}} = 1;~~\lim_{\theta \rightarrow 0}{\frac{1 - \cos \theta}{\theta}} = 0}$$
Here is a geometric proof for the first limit:
When $\theta$ becomes very small, the sector in the figure above looks like:
Imagine what happens to the pictures as $\theta$ ger very small. As $\theta \rightarrow 0$, we see that $\frac{\sin \theta}{\theta} \rightarrow 1$.
What about the second limit involving cosine?
From the figure above we can see that as $\theta \rightarrow 0$, the length $1 - \cos \theta$ of the short segment gets much smaller than the vertical distance $\theta$ along the arc. Hence, $\frac{1 - \cos \theta}{\theta} \rightarrow 0$.
Theorem: Differentiable Implies Continuous¶
If $f$ is differentiable at $x_0$, then $f$ is continuous at $x_0$.
Proof: $\lim_{x \rightarrow x_0} \left( f(x) - f(x_0) \right) = \lim_{x \rightarrow x_0} \left[ \frac{f(x) - f(x_0)}{x - x_0} \right](x - x_0) = f'(x_0) \cdot 0 = 0$
Remember: you can never divide by zero. The first step was to multiply by $\frac{x - x_0}{x - x_0}$. It looks as if this is illegal because when $x = x_0$, we are multiplying by $\frac{0}{0}$. But when computing the limit as $x \rightarrow x_0$ we assume $x \neq x_0$. In other words $x - x_0 \neq 0$. So the proof is valid.