A. What is a derivative?
B. How to differentiate any function you know
Geometric Viewpoint on Derivatives
Tangent Line and Secant Line The derivative is the slope of the line tangent to the graph of $f(x)$. But what is a tangent line, exactly?
Geometric Definition of the Derivative Limit of slopes of secant lines $PQ$ as $Q \rightarrow P$ (with $P$ fixed). The slope of $PQ$: $\text{Slope of PQ} = \frac{\Delta f}{\Delta x} = \frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x}$
$\lim_{\Delta x \to 0} \frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x} = f’(x_0)$
Example 1: $f(x) = \frac{1}{x}$
To find the derivative: $\frac{\Delta f}{\Delta x} = \frac{\frac{1}{x_0 + \Delta x} - \frac{1}{x_0}}{\Delta x}$ $= \frac{x_0 - (x_0 + \Delta x)}{\Delta x \cdot x_0 \cdot (x_0 + \Delta x)} = \frac{-\Delta x}{\Delta x \cdot x_0 \cdot (x_0 + \Delta x)} = -\frac{1}{x_0^2}$ Taking the limit as $\Delta x \rightarrow 0$: $\lim_{\Delta x \to 0} -\frac{1}{x_0^2} = -\frac{1}{x_0^2}$ Thus, $f’(x_0) = -\frac{1}{x_0^2}$
Finding the Tangent Line Using the equation for a line $y - y_0 = f’(x_0)(x - x_0)$: $y - \frac{1}{x_0} = -\frac{1}{x_0^2}(x - x_0)$
Area of the Triangle Formed by the Tangent Line
Notations for the Derivative
Since $y = f(x)$: $\Delta y = \Delta f = f(x) - f(x_0) = f(x_0 + \Delta x) - f(x_0)$ $\frac{\Delta y}{\Delta x} = \frac{\Delta f}{\Delta x}$ Taking the limit as $\Delta x \to 0$: $\frac{\Delta y}{\Delta x} \to \frac{dy}{dx} \quad \text{(Leibniz’ notation)}$ $\frac{\Delta f}{\Delta x} \to f’(x_0) \quad \text{(Newton’s notation)}$
Other notations for the derivative of a function $f$ include: $\frac{df}{dx}, \ f’, \ Df$
Example 2: $f(x) = x^n$ where $n = 1, 2, 3, \ldots$ Using the definition of the difference quotient: $\frac{\Delta y}{\Delta x} = \frac{(x + \Delta x)^n - x^n}{\Delta x}$ Expanding using the binomial theorem: $(x + \Delta x)^n = x^n + nx^{n-1}\Delta x + O((\Delta x)^2)$ $\frac{\Delta y}{\Delta x} = nx^{n-1} + O(\Delta x)$ Taking the limit: $\lim_{\Delta x \to 0} nx^{n-1} = nx^{n-1}$ Thus, $\frac{d}{dx} x^n = nx^{n-1}$
This result extends to polynomials. For example: $\frac{d}{dx} (x^2 + 3x^{10}) = 2x + 30x^9$
Physical Interpretation of Derivatives The derivative represents a rate of change (e.g., speed).
Example: Dropping a pumpkin from a 400-foot building: $y = 400 - 16t^2$ Average speed when the pumpkin hits the ground ($y = 0$): $400 - 16t^2 = 0 \implies t = 5 \ \text{seconds}$ $\text{Average speed} = \frac{400 \ \text{ft}}{5 \ \text{sec}} = 80 \ \text{ft/s}$
Instantaneous velocity at $t = 5$:
$y’ = -32t \implies y’(5) = -160 \ \text{ft/s} \ (\text{about 110 mph})$
$y’$ is negative because the pumpkin is moving downward.
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More about the “rate of change” interpretation of the derivative
$y = f(x)$ $ \Delta x \quad \Delta y $ Figure 1: Graph of a generic function, with $\Delta x$ and $\Delta y$ marked on the graph
$ \frac{\Delta y}{\Delta x} \rightarrow \frac{dy}{dx} \text{ as } \Delta x \rightarrow 0 $
Examples
Sensitivity of measurements An example is carried out on Problem Set 1. In GPS, radio signals give us $h$ up to a certain measurement error. The question is how accurately we can measure $L$.
$\frac{\Delta L}{\Delta h} $
Figures
Limits and Continuity
Easy Limits $\lim_{x \to 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.
$\lim_{\Delta x \to 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 continuous at $x_0$ when: $\lim_{x \to x_0} f(x) = f(x_0) $
Discontinuities
$f(x) = \begin{cases}
x + 1 & x > 0
-x & x \ge 0
\end{cases} $
For $x > 0$: $\lim_{x \to 0} f(x) = 1 $ but $f(0) = 0$. (One can also say $f$ is continuous from the left at 0, not the right.)
Right-hand limit: $\lim_{x \to x_0^+} f(x)$ Left-hand limit: $\lim_{x \to x_0^-} f(x)$
If $\lim_{x \to x_0} f(x)$ exists but is not $f(x_0)$, or if $f(x_0)$ is undefined, the discontinuity is removable.
For example, $\frac{\sin x}{x}$ is defined for $x \neq 0$. We will see later how to evaluate the limit as $x \to 0$.
$\lim_{x \to x_0^-} f(x) \text{ and } \lim_{x \to x_0^+} f(x) $ both exist, but are NOT equal.
$\lim_{x \to 0^+} \frac{1}{x} = \infty $ $\lim_{x \to 0^-} \frac{1}{x} = -\infty $
This function doesn’t even go to $\pm \infty$ — it doesn’t make sense to say it goes to anything. For something like this, we say the limit does not exist.
Picturing the Derivative
Notice 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.
Two Trig Limits Note: In the expressions below, $\theta$ is in radians— NOT degrees!
$\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 $ $\lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta} = 0 $
Geometric Proof for the First Limit:
$\frac{\sin \theta}{\theta} \rightarrow 1 \text{ as } \theta \rightarrow 0 $
Proof for the Second Limit:
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 \to x_0} (f(x) - f(x_0)) = \lim_{x \to x_0} \left( \frac{f(x) - f(x_0)}{x - x_0} \cdot (x - x_0) \right) = 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 always assume $x \neq x_0$. In other words $x - x_0 \neq 0$. So the proof is valid.
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Types of Derivative Formulas
A notational convention we will use today is:
Proof of $(u + v)’ = u’ + v’$ (General) Start by using the definition of the derivative: $(u + v)’(x) = \lim_{\Delta x \to 0} \frac{(u + v)(x + \Delta x) - (u + v)(x)}{\Delta x}$ $= \lim_{\Delta x \to 0} \frac{u(x + \Delta x) + v(x + \Delta x) - u(x) - v(x)}{\Delta x}$ $= \lim_{\Delta x \to 0} \left( \frac{u(x + \Delta x) - u(x)}{\Delta x} + \frac{v(x + \Delta x) - v(x)}{\Delta x} \right)$ $(u + v)’(x) = u’(x) + v’(x)$
Follow the same procedure to prove that $(cu)’ = cu’$.
Derivatives of $\sin x$ and $\cos x$ (Specific) Using trigonometric limits: $\lim_{x \to 0} \frac{\sin x}{x} = 1$
Last time, we computed
\[\lim_{x \to 0} \frac{\sin x}{x} = 1\] \[\left. \frac{d}{dx} (\sin x) \right|_{x=0} = \lim_{\Delta x \to 0} \frac{\sin(0 + \Delta x) - \sin(0)}{\Delta x} = \lim_{\Delta x \to 0} \frac{\sin(\Delta x)}{\Delta x} = 1\] \[\left. \frac{d}{dx} (\cos x) \right|_{x=0} = \lim_{\Delta x \to 0} \frac{\cos(0 + \Delta x) - \cos(0)}{\Delta x} = \lim_{\Delta x \to 0} \frac{\cos(\Delta x) - 1}{\Delta x} = 0\]So, we know the value of $\frac{d}{dx} \sin x$ and $\frac{d}{dx} \cos x$ at $x = 0$. Let us find these for arbitrary $x$.
\[\frac{d}{dx} \sin x = \lim_{\Delta x \to 0} \frac{\sin(x + \Delta x) - \sin(x)}{\Delta x}\]So, we know the value of $\frac{d}{dx} \sin x$ and $\frac{d}{dx} \cos x$ at $x = 0$. Let us find these for arbitrary $x$.
$ \frac{d}{dx} \sin x = \lim_{\Delta x \to 0} \frac{\sin(x + \Delta x) - \sin(x)}{\Delta x} $
Recall: $ \sin(a + b) = \sin(a) \cos(b) + \sin(b) \cos(a) $
So, $ \frac{d}{dx} \sin x = \lim_{\Delta x \to 0} \frac{\sin x \cos \Delta x + \cos x \sin \Delta x - \sin(x)}{\Delta x} $ $ = \lim_{\Delta x \to 0} \left[ \sin x \left( \frac{\cos \Delta x - 1}{\Delta x} \right) + \cos x \left( \frac{\sin \Delta x}{\Delta x} \right) \right] $ $ = \lim_{\Delta x \to 0} \sin x \left( \frac{\cos \Delta x - 1}{\Delta x} \right) + \lim_{\Delta x \to 0} \cos x \left( \frac{\sin \Delta x}{\Delta x} \right) $
Since $ \frac{\cos \Delta x - 1}{\Delta x} \to 0 $ and $ \frac{\sin \Delta x}{\Delta x} \to 1 $ the equation above simplifies to: $ \frac{d}{dx} \sin x = \cos x $
A similar calculation gives: $ \frac{d}{dx} \cos x = -\sin x $
Product Formula (General) $(uv)’ = u’v + uv’$
Proof: $(uv)’ = \lim_{\Delta x \to 0} \frac{(uv)(x + \Delta x) - (uv)(x)}{\Delta x}$ $= \lim_{\Delta x \to 0} \frac{u(x + \Delta x)v(x + \Delta x) - u(x)v(x)}{\Delta x}$
By adding and subtracting $u(x + \Delta x)v(x)$: $(uv)’ = \lim_{\Delta x \to 0} \frac{u(x + \Delta x)v(x + \Delta x) - u(x + \Delta x)v(x) + u(x + \Delta x)v(x) - u(x)v(x)}{\Delta x}$ $= \lim_{\Delta x \to 0} \left( \frac{u(x + \Delta x) - u(x)}{\Delta x}\right) v(x) + u(x + \Delta x) \left( \frac{v(x + \Delta x) - v(x)}{\Delta x} \right)$
The limit of a sum is the sum of the limits. \(\left[ \lim_{\Delta x \to 0} \frac{u(x + \Delta x) - u(x)}{\Delta x} \right] v(x) + \lim_{\Delta x \to 0} \left( u(x + \Delta x) \left[ \frac{v(x + \Delta x) - v(x)}{\Delta x} \right] \right)\)
$= u’(x)v(x) + u(x)v’(x)$
This proof assumes that $u$ and $v$ have derivatives, which implies both functions are continuous.
Quotient Formula (General) To calculate the derivative of $\frac{u}{v}$, use the same notations for $\Delta u$ and $\Delta v$:
$\frac{u(x + \Delta x)}{v(x + \Delta x)} - \frac{u(x)}{v(x)} = \frac{(u + \Delta u)v - u(v + \Delta v)}{(v + \Delta v)v} = \frac{(\Delta u)v - u(\Delta v)}{(v + \Delta v)v}$
Divide by $\Delta x$ and take the limit as $\Delta x \to 0$: $\frac{\Delta u}{\Delta x} v - u \frac{\Delta v}{\Delta x} = \frac{u’v - uv’}{v^2}$
Therefore, $\left( \frac{u}{v} \right)’ = \frac{u’v - uv’}{v^2}$
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Chain Rule We’ve covered differentiation procedures for addition, subtraction, and multiplication. Now, we’ll discuss the differentiation of compositions of functions.
Example 1: Given: $y = f(x) = \sin x, \quad x = g(t) = t^2$ So, $y = f(g(t)) = \sin(t^2)$
To find $\frac{dy}{dt}$: $t = t_0 + \Delta t$ $x = x_0 + \Delta x$ $y = y_0 + \Delta y$ $\frac{\Delta y}{\Delta t} = \frac{\Delta y}{\Delta x} \cdot \frac{\Delta x}{\Delta t}$
As $\Delta t \rightarrow 0$, $\Delta x \rightarrow 0$ due to continuity, thus: $\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$
In this example: $\frac{dx}{dt} = 2t$ $\frac{dy}{dx} = \cos x$
So, $\frac{d}{dt} \left( \sin(t^2) \right) = \frac{dy}{dx} \cdot \frac{dx}{dt} = \cos(t^2) \cdot 2t$
Another notation for the chain rule: $\frac{d}{dt} f(g(t)) = f’(g(t)) \cdot g’(t)$ or $\frac{d}{dx} f(g(x)) = f’(g(x)) \cdot g’(x)$
Example 1 (continued): Composition of functions: $f(x) = \sin x$ $g(x) = x^2$ $(f \circ g)(x) = f(g(x)) = \sin(x^2)$ $(g \circ f)(x) = g(f(x)) = \sin^2(x)$
Note: $f \circ g \neq g \circ f$. Composition is not commutative.
Example 2: $\frac{d}{dx} \cos \left( \frac{1}{x} \right)$
Let $u = \frac{1}{x}$, $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$
$\frac{dy}{du} = -\sin(u)$ $\frac{du}{dx} = -\frac{1}{x^2}$
So, $\frac{d}{dx} \cos \left( \frac{1}{x} \right) = -\sin \left( \frac{1}{x} \right) \cdot \left( -\frac{1}{x^2} \right) = \frac{\sin \left( \frac{1}{x} \right)}{x^2}$
Example 3: $\frac{d}{dx} x^{-n}$
There are two ways to proceed: $x^{-n} = \frac{1}{x^n}$
Using the power rule: $\frac{d}{dx} x^{-n} = \frac{d}{dx} \left( \frac{1}{x^n} \right) = -nx^{-n-1}$
Using the chain rule: $x^{-n} = \left( x^n \right)^{-1}$ $\frac{d}{dx} x^{-n} = \frac{d}{dx} \left( \left( x^n \right)^{-1} \right) = -n x^{-n-1}$
Higher Derivatives Higher derivatives are derivatives of derivatives. For instance, if $g = f’$, then $h = g’$ is the second derivative of $f$. We write $h = (f’)’ = f’’$.
Notations:
Higher derivatives are straightforward: keep taking the derivative.
Example: $D^n x = ?$
Start small and look for a pattern: $D x = 1$ $D^2 x = D (2x) = 2$ $D^3 x = D^2 (3x^2) = D (6x) = 6$ $D^4 x = D^3 (4x^3) = D^2 (12x^2) = D (24x) = 24$
So, $D^n x^n = n!$
The notation $n!$ (n factorial) is defined by $n! = n (n-1) \cdots 2 \cdot 1$.
Proof by Induction: We’ve checked the base case ($n = 1$).
Induction step: Suppose we know $D^n x^n = n!$ (n-th case). Show it holds for the (n+1)-st case: $D^{n+1} x^{n+1} = D^n (D x^{n+1}) = D^n ((n+1) x^n) = (n+1) D^n x^n = (n+1)(n!)$ So, $D^{n+1} x^{n+1} = (n+1)!$
Proved!
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Implicit Differentiation
Example 1: To differentiate $x^a$: $\frac{d}{dx} (x^a) = ax^{a-1}$
We proved this for $a = 0, 1, 2, \ldots$ and extended it to $a = -1, -2, \ldots$. Let’s extend this to rational numbers as well:
$y = x^{m/n}$ To compute $\frac{dy}{dx}$: $y^n = x^m$ $n y^{n-1} \frac{dy}{dx} = m x^{m-1}$ $\frac{dy}{dx} = \frac{m x^{m-1}}{n y^{n-1}}$
Since $y = x^{m/n}$: $\frac{dy}{dx} = \frac{m x^{m-1}}{n x^{(m/n)(n-1)}} = \frac{m x^{m-1}}{n x^{m - m/n}} = \frac{m x^{m-1}}{n x^{m - m/n}} = \frac{m}{n} x^{m/n - 1}$
This is the same result we hoped to get.
Example 2: Equation of a circle with radius 1: $x^2 + y^2 = 1$ $y^2 = 1 - x^2$ $y = \pm \sqrt{1 - x^2}$
For the positive case: $y = \sqrt{1 - x^2}$ $\frac{dy}{dx} = \frac{1}{2}(1 - x^2)^{-1/2}(-2x) = -\frac{x}{\sqrt{1 - x^2}}$
Using implicit differentiation: $x^2 + y^2 = 1$ Differentiating both sides: $2x + 2y \frac{dy}{dx} = 0$ $2y \frac{dy}{dx} = -2x$ $\frac{dy}{dx} = -\frac{x}{y}$
Same answer!
Example 3: $y^3 + xy^2 + 1 = 0$
Implicit differentiation: $3y^2 \frac{dy}{dx} + y^2 + 2xy \frac{dy}{dx} = 0$ Solving for $\frac{dy}{dx}$: $\frac{dy}{dx} (3y^2 + 2xy) = -y^2$ $\frac{dy}{dx} = \frac{-y^2}{3y^2 + 2xy}$
Inverse Functions If $y = f(x)$ and $g(y) = x$, then $g$ is the inverse function of $f$, denoted $f^{-1}$: $x = g(y) = f^{-1}(y)$
Using implicit differentiation to find the derivative of the inverse function: $y = f(x)$ $f^{-1}(y) = x$ Differentiating both sides: $\frac{d}{dx} (f^{-1}(y)) = \frac{d}{dx} (x) = 1$ By the chain rule: $\frac{d}{dy} (f^{-1}(y)) \frac{dy}{dx} = 1$ So, $\frac{d}{dy} (f^{-1}(y)) = \frac{1}{\frac{dx}{dy}}$
Example: $y = \arctan(x)$ $\tan(y) = x$ Differentiating both sides: $\frac{d}{dx} [\tan(y)] = \frac{d}{dx} (x) = 1$ $\sec^2(y) \frac{dy}{dx} = 1$ $\frac{dy}{dx} = \cos^2(y)$
Using geometry to simplify: $\tan(y) = x$ $\sec^2(y) = 1 + \tan^2(y) = 1 + x^2$ $\cos^2(y) = \frac{1}{1 + x^2}$
Thus, $\frac{dy}{dx} = \frac{1}{1 + x^2}$
Graphing an Inverse Function If $y = f(x)$ and $g(y) = f^{-1}(y) = x$, to graph $g$ and $f$ together, we write $g$ as a function of $x$. Swapping the variables $x$ and $y$ illustrates the inverse relationship: $f^{-1}(f(x)) = x$ $f(f^{-1}(x)) = x$
Visualize $f^{-1}$ as the graph of $f$ reflected about the line $y = x$.
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Taking the Derivatives of Exponentials and Logarithms
Background We always assume the base $a$ is greater than 1. $a^0 = 1; \quad a^1 = a; \quad a^2 = a \cdot a$ $a^{x_1 + x_2} = a^{x_1} a^{x_2}$ $(a^{x_1})^{x_2} = a^{x_1 x_2}$ $a^{\frac{p}{q}} = \sqrt[q]{a^p} \quad \text{(where p and q are integers)}$
To define $a^r$ for real numbers $r$, fill in by continuity.
Today’s Main Task: Find $\frac{d}{dx} a^x$
We can write: $\frac{d}{dx} a^x = \lim_{\Delta x \to 0} \frac{a^{x + \Delta x} - a^x}{\Delta x}$
We can factor out $a^x$: $\frac{d}{dx} a^x = a^x \lim_{\Delta x \to 0} \frac{a^{\Delta x} - 1}{\Delta x}$
Let’s call: $M(a) \equiv \lim_{\Delta x \to 0} \frac{a^{\Delta x} - 1}{\Delta x}$
We don’t yet know what $M(a)$ is, but we can say: $\frac{d}{dx} a^x = M(a) a^x$
Two Descriptions of $M(a)$
Indeed, $M(a) = \lim_{\Delta x \to 0} \frac{a^{0 + \Delta x} - a^0}{\Delta x} = \left. \frac{d}{dx} a^x \right|_{x=0}$
The trick to figuring out $M(a)$ is to define $e$ as the number such that $M(e) = 1$.
We can estimate the slope $M(a)$ for $a = 2$ and $a = 4$:
Somewhere between 2 and 4, there is a base whose slope at $x = 0$ is 1.
Thus, we define $e$ as the unique number such that: $M(e) = 1$ $\lim_{h \to 0} \frac{e^h - 1}{h} = 1$ $\left. \frac{d}{dx} e^x \right|_{x=0} = 1$
So: $\frac{d}{dx} e^x = e^x$
Natural Logarithm (Inverse Function of $e^x$) To understand $M(a)$ better, we study the natural log function $\ln(x)$. This function is defined as follows:
Note:
Use implicit differentiation to find $\frac{d}{dx} \ln(x)$: $w = \ln(x)$ $e^w = x$ $\frac{d}{dx} (e^w) = \frac{d}{dx} (x)$ $e^w \frac{dw}{dx} = 1$ $\frac{dw}{dx} = \frac{1}{e^w} = \frac{1}{x}$
So: $\frac{d}{dx} \ln(x) = \frac{1}{x}$
General Exponential Functions Finally, what about $\frac{d}{dx} a^x$?
Method 1: Write in Base $e$ and Use Chain Rule Rewrite $a$ as $e^{\ln(a)}$: $a^x = (e^{\ln(a)})^x = e^{x \ln(a)}$
Using the chain rule: $\frac{d}{dx} e^{x \ln(a)} = \ln(a) e^{x \ln(a)} = \ln(a) a^x$
Recall: $\frac{d}{dx} a^x = M(a) a^x$ So $M(a) = \ln(a)$.
Even if starting with another base like 10, the natural logarithm appears: $\frac{d}{dx} 10^x = (\ln 10) 10^x$
Method 2: Logarithmic Differentiation The idea is to find $f(x)$ by finding $\ln(f(x))$: $u = f(x)$ $\frac{d}{dx} \ln(u) = \frac{1}{u} \frac{du}{dx}$ $\frac{du}{dx} = u \frac{d}{dx} \ln(u)$
Apply this to $f(x) = a^x$: $\ln(f(x)) = x \ln(a)$ $\frac{d}{dx} \ln(f(x)) = \ln(a)$
So: $\frac{du}{dx} = f(x) \ln(a) = a^x \ln(a)$
Example 1: $\frac{d}{dx} (x^x)$
Use logarithmic differentiation: $f = x^x$ $\ln(f) = x \ln(x)$
Differentiate: $\frac{d}{dx} \ln(f) = \frac{d}{dx} (x \ln(x)) = \ln(x) + 1$
So: $\frac{du}{dx} = x^x (\ln(x) + 1)$
Example 2: Evaluate $\lim_{k \to \infty} \left(1 + \frac{1}{k}\right)^k$
Find the limit of the logarithm: $\lim_{k \to \infty} \ln \left(1 + \frac{1}{k}\right)^k = \lim_{k \to \infty} k \ln \left(1 + \frac{1}{k}\right)$
This expression has two parts:
So: $\lim_{k \to \infty} k \ln \left(1 + \frac{1}{k}\right) = \lim_{h \to 0} \frac{\ln(1 + h)}{h} = \left. \frac{d}{dx} \ln(x) \right|_{x=1} = 1$
Thus: $\lim_{k \to \infty} \left(1 + \frac{1}{k}\right)^k = e$
Remarks: