[Calculus Series Part 2] Week 1: Limits of Exponential, Log, and Trig Functions

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In the first week of calculus we step into the world of transcendental functions: functions such as exponentials, logarithms, and trigonometric functions that go beyond ordinary polynomial expressions. Beyond polynomials, we now meet exponential, logarithmic, and trigonometric functions in full. The goal this week is not to memorize a bag of formulas, but to follow one connected storyline: why ee is special, why the derivative of lnx\ln x becomes 1/x1/x, and why the limit sinx/x\sin x/x unlocks trig differentiation.

The series context also matters. In Part 1 we reviewed limits as the language of calculus. In this post we use that same limit idea to build the derivative formulas for new families of functions, and next week we will use those results to differentiate sine and cosine directly.

1. Lesson 1: Limits and derivatives of exponentials and logarithms

1.1 Why the natural base e?

One of the most important constants in calculus is the natural base ee. The driving question is simple: can we choose a base aa so that the derivative of axa^x equals axa^x itself?

Start from the derivative definition:

ddxax=limh0ax+haxh=limh0axahaxh=axlimh0ah1h\frac{d}{dx} a^x = \lim_{h \to 0} \frac{a^{x+h}-a^x}{h} = \lim_{h \to 0} \frac{a^x a^h-a^x}{h} = a^x \lim_{h \to 0} \frac{a^h - 1}{h}

The important point is that the factor

limh0ah1h\lim_{h \to 0} \frac{a^h - 1}{h}

depends only on the base aa, not on xx. So the derivative of every exponential has the form "the same function times a constant." For example,

ddx2x=2xln20.6932x.\frac{d}{dx} 2^x = 2^x \ln 2 \approx 0.693 \cdot 2^x.

The unique base that makes that constant exactly 11 is ee, so

ddxex=ex.\frac{d}{dx} e^x = e^x.

In other words, exe^x is the unique exponential whose rate of change at each point equals its value at that point, which is why it appears everywhere in calculus.

When a phenomenon grows proportionally to its current size—population growth, continuous compounding, radioactive decay, solutions of differential equations—it inevitably looks like exe^x or a scaled version of it. The constant ee is therefore not a convenience—it acts like a fundamental unit of continuous change.

1.2 Definitions of e and logarithmic differentiation

There are many equivalent ways to define ee, but courses usually present two at once.

e=limn(1+1n)ne = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n

This is the limit of continuous compounding. If you compound interest more and more frequently-semiannually, quarterly, monthly, and ultimately infinitely often-you naturally arrive at that limit. This limit converges to approximately 2.718282.71828.

Another view comes from the differential equation y=yy' = y with y(0)=1y(0)=1; the base that satisfies this condition becomes ee. These two viewpoints describe the same number. The compounding formula gives an intuitive limit model, while the differential-equation view explains why exe^x is the natural exponential for calculus.

ddxex=ex,ddxax=axlna\frac{d}{dx} e^x = e^x, \qquad \frac{d}{dx} a^x = a^x \ln a

Now connect this to logarithms. Since lnx\ln x is the inverse of exe^x, let

y=lnx.y = \ln x.

Then ey=xe^y = x. Differentiate both sides with respect to xx:

eydydx=1.e^y \frac{dy}{dx} = 1.

Because ey=xe^y = x, this becomes

dydx=1ey=1x.\frac{dy}{dx} = \frac{1}{e^y} = \frac{1}{x}.

So

ddxlnx=1x,ddxlogax=1xlna\frac{d}{dx} \ln x = \frac{1}{x}, \qquad \frac{d}{dx} \log_a x = \frac{1}{x \ln a}

Here lnx\ln x is special because it is the inverse of exe^x. If exe^x is the most natural exponential, then lnx\ln x is the most natural logarithm. Common logarithms (base 10) or binary logarithms (base 2) are useful, but derivatives are simplest with lnx\ln x.

Keep these correspondences in mind:

  • Exponentials model repeated multiplicative growth, such as compounding by the same percentage.
  • Logarithms turn products into sums, which is why they simplify multiplication-based relationships.
  • Derivatives look simplest with exe^x and lnx\ln x, because no extra conversion factor is needed.

1.3 Quick computation tour

See how the formulas work immediately:

ddxe3x=3e3x,\frac{d}{dx} e^{3x} = 3e^{3x}, ddxln(2x+1)=22x+1,\frac{d}{dx} \ln(2x+1) = \frac{2}{2x+1}, ddxlog2x=1xln2.\frac{d}{dx} \log_2 x = \frac{1}{x \ln 2}.

The first combines an exponential derivative with the chain rule; the second pairs logarithmic differentiation with the chain rule; the third reminds us that every non-natural base injects a lna\ln a factor below.

💻 Python snippet (NumPy)

n_values = np.array([1, 10, 100, 1000, 10000, 100000])
e_approx = (1 + 1 / n_values) ** n_values

print("n\t(1+1/n)^n")
print("-" * 25)
for n, e_val in zip(n_values, e_approx):
    print(f"{n}\t{e_val:.10f}")

print(f"\nActual e: {np.e:.10f}")

1.4 Trig addition formulas and a key limit

Addition formulas for sine and cosine are the doors to later derivative formulas. When we differentiate sinx\sin x from first principles, sin(x+h)\sin(x+h) appears, so we need a way to expand that expression.

sin(A+B)=sinAcosB+cosAsinB\sin(A + B) = \sin A \cos B + \cos A \sin B cos(A+B)=cosAcosBsinAsinB\cos(A + B) = \cos A \cos B - \sin A \sin B

Visualize them as rotations on the unit circle: the formulas are not arbitrary, they express how coordinates change under rotation. We need this structure so we can expand x+hx+h in terms of xx and hh and plug it into limit definitions.

From these formulas we derive the most important limit in this unit:

limx0sinxx=1.\lim_{x \to 0} \frac{\sin x}{x} = 1.

Seeing how sinx/x\sin x / x approaches 1 as x0x \to 0 (in radians) is essential because this limit becomes the starting point of trig differentiation.

1.5 Proving the fundamental limit

The crucial part is not the answer itself but the reasoning for

limx0sinxx=1.\lim_{x \to 0} \frac{\sin x}{x} = 1.

We need this limit to justify trig derivatives from the definition. The standard proof uses the unit circle plus the squeeze theorem. The core geometric idea is simple: on the unit circle, the straight chord is shorter than the arc, and the arc is shorter than the tangent segment built outside the circle.

For 0<x<π20<x<\frac{\pi}{2}, compare three regions with central angle xx:

  • An inscribed triangle with area 12sinx\frac{1}{2}\sin x
  • A sector with area 12x\frac{1}{2}x because sector area on the unit circle is 12r2θ=12x\frac{1}{2}r^2\theta = \frac{1}{2}x
  • A larger triangle formed with the tangent line, with area 12tanx\frac{1}{2}\tan x

Therefore,

12sinx<12x<12tanx,\frac{1}{2}\sin x < \frac{1}{2}x < \frac{1}{2}\tan x,

and multiplying by 2 gives

sinx<x<tanx.\sin x < x < \tan x.

Using tanx=sinxcosx\tan x = \frac{\sin x}{\cos x},

sinx<x<sinxcosx.\sin x < x < \frac{\sin x}{\cos x}.

Since sinx>0\sin x > 0 on that interval, divide all terms by sinx\sin x:

1<xsinx<1cosx.1 < \frac{x}{\sin x} < \frac{1}{\cos x}.

All three quantities are positive, so taking reciprocals reverses the inequalities and gives

cosx<sinxx<1.\cos x < \frac{\sin x}{x} < 1.

As x0x \to 0, cosx1\cos x \to 1, and the squeeze theorem yields

limx0sinxx=1.\lim_{x \to 0} \frac{\sin x}{x} = 1.

We performed the proof for x>0x>0, but sin(x)=sinx\sin(-x)=-\sin x shows that the same value holds from the left, so the two-sided limit exists.

This result gives us two other limits immediately:

limx0tanxx=limx0sinxx1cosx=1,\lim_{x \to 0} \frac{\tan x}{x} = \lim_{x \to 0} \frac{\sin x}{x} \cdot \frac{1}{\cos x} = 1,

and

1cosx=(1cosx)(1+cosx)1+cosx=1cos2x1+cosx=sin2x1+cosx,1-\cos x = \frac{(1-\cos x)(1+\cos x)}{1+\cos x} = \frac{1-\cos^2 x}{1+\cos x} = \frac{\sin^2 x}{1+\cos x},

so

1cosxx=sinxxsinx1+cosxx00.\frac{1-\cos x}{x} = \frac{\sin x}{x}\cdot\frac{\sin x}{1+\cos x} \xrightarrow[x \to 0]{} 0.

These two limits reappear directly in next week's derivations of (sinx)=cosx(\sin x)'=\cos x and (cosx)=sinx(\cos x)'=-\sin x. Remember one warning: these limits equal the stated values only in radians. If you measure xx in degrees, the numbers differ. In fact,

limx0sin(x)x=π180,\lim_{x \to 0} \frac{\sin(x^\circ)}{x} = \frac{\pi}{180},

not 11. Radians keep arc length and angle linked, which is why trig derivatives stay clean.

It is also useful to keep the following together:

limx01cosxx=0,limx0tanxx=1.\lim_{x \to 0} \frac{1-\cos x}{x} = 0, \qquad \lim_{x \to 0} \frac{\tan x}{x} = 1.

For the second limit we already used

tanxx=sinxx1cosx,\frac{\tan x}{x} = \frac{\sin x}{x} \cdot \frac{1}{\cos x},

and the first limit is the auxiliary piece we need when differentiating cosine.

2. Seeing this week as one stream

The week’s ideas are tightly linked:

  1. Exploring exponential derivatives reveals the special base ee.
  2. Its inverse lnx\ln x provides the simplest logarithmic derivative.
  3. In trigonometry, the addition formulas plus key limits prepare us to differentiate sine and cosine.

So this week is less about "introducing new functions" and more about gathering the raw materials needed to build derivative formulas.

3. Weekly checkpoints

Focus on connecting ee, exponential/log derivatives, trig addition formulas, and the limit limx0sinxx\lim_{x \to 0} \frac{\sin x}{x}.

Checkpoint 1

Find the derivative of f(x)=e2xlnxf(x) = e^{2x} \ln x.

Show answer

Apply the product rule and chain rule:

f(x)=2e2xlnx+e2x1x=e2x(2lnx+1x).f'(x) = 2e^{2x} \ln x + e^{2x} \cdot \frac{1}{x} = e^{2x}\left(2\ln x + \frac{1}{x}\right).

Checkpoint 2

Compute limx0tanxx\lim_{x \to 0} \frac{\tan x}{x}.

Show answer
limx0tanxx=limx0sinxx1cosx=1.\lim_{x \to 0} \frac{\tan x}{x} = \lim_{x \to 0} \frac{\sin x}{x} \cdot \frac{1}{\cos x} = 1.

Checkpoint 3

Use the key limit to find limx0sin(3x)x\lim_{x \to 0} \frac{\sin(3x)}{x}.

Show answer
limx0sin(3x)x=limx0sin(3x)3x3=13=3.\lim_{x \to 0} \frac{\sin(3x)}{x} = \lim_{x \to 0} \frac{\sin(3x)}{3x} \cdot 3 = 1 \cdot 3 = 3.

Checkpoint 4

Differentiate f(x)=ex+lnxcosxf(x) = e^x + \ln x - \cos x.

Show answer

Differentiate term by term:

f(x)=ex+1x+sinx.f'(x) = e^x + \frac{1}{x} + \sin x.

Checkpoint 5

Explain in your own words why limx0sinxx=1\lim_{x \to 0} \frac{\sin x}{x} = 1 launches trig differentiation.

Hint

When you write (sinx)(\sin x)' using the limit definition, sin(x+h)\sin(x+h) appears. After expanding with the addition formula, the terms sinh/h\sin h / h and (cosh1)/h(\cos h - 1)/h show up; knowing their limits is the only way to finish the derivation and get cosx\cos x.

4. Frequent pitfalls

  • Forgetting the lna\ln a factor and assuming ddxax=ax\frac{d}{dx} a^x = a^x.
  • Dropping the lna\ln a in the denominator when differentiating logax\log_a x.
  • Applying limx0sinxx=1\lim_{x \to 0} \frac{\sin x}{x} = 1 to degree measure instead of radians.
  • Ignoring that lnx\ln x is defined only for x>0x>0 and pretending it differentiates over all reals.

5. Next week

Next week dives into trig differentiation and the chain rule properly. We will connect the addition formulas and limits from this week to explicit derivative formulas.

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