Limits, Continuity, Differentiability and Applications of Derivatives

1Standard limits and leading forms

A limit is the value a function approaches, not the value it attains. Three forms earn their keep on every paper: \lim_{x\to 0}\dfrac{\sin x}{x}=1, \lim_{x\to 0}\dfrac{e^x-1}{x}=1 and \lim_{x\to 0}(1+x)^{1/x}=e. For a \tfrac{0}{0} trig or exponential limit, replace each factor by its leading term (\sin x\approx x, e^x-1\approx x, \ln(1+x)\approx x) before differentiating twice.

Imagine the factor \sin\theta/\theta flattening to 1 as \theta\to 0; the work on the page is getting that factor to appear, not sketching the flat line.

Rewriting into a standard form

1. Name the indeterminate lookIf direct substitution gives \tfrac{0}{0} or \tfrac{\infty}{\infty}, do not stop — rewrite.
2. Reach a standard factorUse an identity or a leading-term swap until a factor like \dfrac{\sin\theta}{\theta} appears.
3. Take the known limitEach standard factor contributes its listed value; multiply the constants that remain.

2Indeterminate forms and L'Hôpital

The shapes \tfrac{0}{0} and \tfrac{\infty}{\infty} are invitations to rewrite, not answers. L'Hôpital's rule says that when those forms appear and the derivatives exist, \lim\dfrac{f}{g}=\lim\dfrac{f'}{g'} — provided the second limit exists. A 1^\infty form is different: convert with \lim f^g = e^{\lim g(f-1)} and finish as an ordinary product limit.

Figure. 0/0 and \infty/\infty rewrite by L'Hopital to f'/g' when that limit exists. A 1^\infty form is different: convert first to e^{\lim g(f-1)} and finish the exponent as an ordinary product limit.

Handling a 1^\infty form

1. Recognise the shapeBase \to 1 and exponent \to\infty (or -\infty) together.
2. Pass to the exponentialWrite \lim f^g = e^{\lim g\ln f}, or the fast form e^{\lim g(f-1)} when f\to 1.
3. Finish the exponentThe exponent is now a \tfrac{0}{0} or product limit; evaluate it, then exponentiate.

3Continuity at a point

f is continuous at a when the limit exists and equals the value: \lim_{x\to a}f(x)=f(a). That single equation packs three checks — left-hand limit, right-hand limit, and f(a) itself — and a jump, a hole or a value sitting off the limit fails continuity for different reasons.

Figure. A jump at 0: left limit, right limit and f(0) are not one number, so the three-way continuity test fails.

4Differentiability and the cusp

Differentiability at a needs the difference quotient to approach one number from both sides, so the left-hand and right-hand derivatives must agree. Differentiability implies continuity, but continuity does not imply differentiability — the cusp of |x| at 0 is continuous and still has left derivative -1 against right derivative +1.

Figure. Two straight rays meet at the origin with slopes -1 and +1. Continuity only needs the graph to join; differentiability needs one tangent, and these rays refuse to share one.

Testing differentiability at a kink

1. Form both difference quotientsCompute \lim_{h\to 0^-}\dfrac{f(a+h)-f(a)}{h} and the matching right-hand limit.
2. Compare the two numbersEqual values give f'(a); unequal values mean a corner or cusp and no derivative.
3. Do not stop at continuityA continuous graph can still bend sharply enough that no single tangent exists.

5Product, quotient and chain rules

Composite and product expressions are dismantled by three rules: (uv)'=u'v+uv', \left(\dfrac{u}{v}\right)'=\dfrac{u'v-uv'}{v^2}, and the chain rule \dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}. Standard pieces include \dfrac{d}{dx}\sin x=\cos x and \dfrac{d}{dx}\ln x=\dfrac{1}{x}; the rules say how those pieces combine.

No figure: keep the product, quotient and chain formulas on one sheet and practise naming outer versus inner.

Differentiating a composite

1. Name the outer and innerWrite y=f(u) with u=g(x) so the chain rule has two clear factors.
2. Differentiate each layerCompute f'(u) and g'(x) with the standard derivatives.
3. MultiplyReturn \dfrac{dy}{dx}=f'(u)\,g'(x), then substitute u back in x.

6Monotonicity and local extrema

f increases on intervals where f'(x)>0 and decreases where f'(x)<0. Critical points satisfy f'(x)=0 (or f' fails to exist). The second-derivative test reads the bend: f''(c)>0 is a local minimum, f''(c)<0 a local maximum. On a closed interval the absolute extreme may sit at an endpoint instead — compare critical values with both endpoints.

Figure. Area against half-width rises from the degenerate thin rectangles, peaks at x=2 with A=32, then falls back to zero as the parabola meets the axis. The peak is where A'=0 and A''<0.

Maximising a single-variable quantity

1. Write one functionExpress the quantity to optimise in a single variable, with a stated domain.
2. Solve f'=0Find critical points inside the domain.
3. Confirm and compareUse f'' for a local test; on a closed interval also evaluate the endpoints.

7Tangents, normals and related rates

The tangent slope at x_0 is f'(x_0); the normal is perpendicular, so its slope is -1/f'(x_0) when the tangent slope is nonzero. Related-rate problems never differentiate with respect to the wrong variable: if y depends on x and both change with t, the chain rule gives \dfrac{dy}{dt}=\dfrac{dy}{dx}\dfrac{dx}{dt}.

Figure. The dashed line is the tangent to y=x^2 at x=1, where f'=2. The normal would be the perpendicular through the same point, with slope -1/2, not drawn here.

Reading a related-rate link

1. Name the dependenceWrite the geometric or algebraic relation between the changing quantities.
2. Differentiate with respect to tApply \dfrac{dy}{dt}=\dfrac{dy}{dx}\dfrac{dx}{dt} (or the multivariate analogue).
3. Substitute the known rateInsert the given dx/dt and the current x, then solve for the unknown rate.