Limits, Continuity and Differentiability

Continuity of piecewise functions

A function f is continuous at a point a when the left-hand limit \lim_{x\to a^-}f(x), the right-hand limit \lim_{x\to a^+}f(x), and the value f(a) are all equal. For a piecewise function, only the junction points where the defining formula changes need checking. At each junction both one-sided limits must equal the value assigned there. For example, if f(x)=ax+1 for x<2 and f(x)=3x-b for x\geq 2 is continuous, then 2a+1=6-b. Writing such equalities gives a system of equations in the unknown constants, which is why many questions reduce to solving two linear equations simultaneously.

Existence of a limit is a weaker requirement than continuity. The limit at a exists exactly when the two one-sided limits are equal, and this common value need not equal f(a). When only this equality is imposed, the resulting equation may be a quadratic in the unknown constant. Every root must then be tested against the conditions of the problem, including whether a denominator becomes zero.

An expression of the form \frac{x-c}{|x-c|} equals -1 for x<c and +1 for x>c, so it produces a jump at c. The greatest integer function [x] is constant on each interval between consecutive integers and takes the lower integer as its value. Thus [x]=0 on (0,1) and [x]=-1 on (-1,0). Composite expressions built from such functions have constant one-sided behaviour near a point, so each one-sided limit can be read off directly. A function built from [x] is continuous on each open interval where it is constant, and it may have jumps at integers.

Differentiability at a junction is a stronger requirement. It needs continuity together with equal left and right derivatives, so the value and the slope of the two pieces must match at the junction.

A polynomial piece such as ax(x+1)+b has limit at x=0 found by substitution. Every term containing a factor x vanishes, leaving only the constant term b.

An exponential piece c\cdot b^{x} with b>0 has limit c as x\to 0, because b^{0}=1. A continuity condition at 0 therefore becomes a simple equation in c and the other constants.

Chain rule differentiation

The chain rule states that if y=F(g(x)) then \frac{dy}{dx}=F'(g(x))\,g'(x). The outer derivative is evaluated at the inner function, and the result is multiplied by the inner derivative. Omitting the inner factor is the most common error. For instance, if h(x)=F(x^2) then h'(x)=2x\,F'(x^2).

The same rule applies to trigonometric and logarithmic compositions. We have \frac{d}{dx}\sin(\ln x)=\frac{\cos(\ln x)}{x}, and more generally \frac{d}{dx}\ln|u|=\frac{u'}{u}. A sum of sine and cosine of \ln x therefore carries a factor \frac{1}{x}, and its value at a point is obtained by substituting into the simplified derivative.

The derivative of one function with respect to another is a ratio, \frac{du}{dv}=\frac{du/dx}{dv/dx}, in which the common variable x cancels. If u=e^x is treated as the new variable, then \frac{d}{du}e^{u}=e^{u}, so the derivative of e^{e^x} with respect to e^x is e^{e^x} itself. The ratio of derivatives can also be a constant, as when \sin^2 x is differentiated with respect to \cos^2 x, where both derivatives are multiples of \sin 2x.

When a function contains |\ln x|, its derivative depends on the sign of \ln x on a neighbourhood of the point. For 0<x<1 we have \ln x<0, so |\ln x|=-\ln x and its derivative is -\frac{1}{x}. For x>1 the expression is positive, so the derivative is \frac{1}{x}.

A piecewise function composed with itself must be simplified interval by interval. On (1,2) the inner value \ln x lies in (0,1), where the absolute value changes the sign. The outer function then gives -\ln(\ln x), and the chain rule yields -\frac{1}{x\ln x}.

Angles given in degrees must be converted to radians before differentiating trigonometric functions. Since x^\circ=\frac{\pi x}{180}, the chain rule contributes the factor \frac{\pi}{180}. For example, \frac{d}{dx}\operatorname{cosec}(x^\circ)=-\frac{\pi}{180}\operatorname{cosec}(x^\circ)\cot(x^\circ).

For a base a>0, \frac{d}{dx}a^{x}=a^{x}\ln a. A piece of the form c\cdot 5^{x} therefore has derivative c\cdot 5^{x}\ln 5, and substituting the point gives a value containing \ln 5.

The chain rule runs in reverse for integrals whose numerator combines the denominator D and its derivative D'. If N=AD+BD', then \int\frac{N}{D}\,dx=Ax+B\ln|D|+c, because \frac{d}{dx}\ln|D|=\frac{D'}{D}. The constants A and B are found by comparing coefficients of the trigonometric terms, and A gives the coefficient of x while B gives the coefficient of the logarithm.

A denominator containing a difference of square roots is simplified by multiplying by the conjugate, since (\sqrt{u}-\sqrt{v})(\sqrt{u}+\sqrt{v})=u-v. After this step the integrand is a sum of powers. Because \int(x+1)^{1/2}\,dx=\frac{2}{3}(x+1)^{3/2}+c, the reverse chain rule with inside derivative 1 supplies the coefficient.

Limit definition of derivative

The derivative at a is defined by f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}. Any limit with this difference-quotient form is therefore a derivative of the function at that point and can be found by differentiating. For \sin^2 x, the limit of \frac{\sin^2(x+h)-\sin^2 x}{h} as h\to 0 is 2\sin x\cos x=\sin 2x by the double-angle identity.

When f(0)=0, the quotient \frac{f(x)}{x} equals \frac{f(x)-f(0)}{x-0}, so \lim_{x\to 0}\frac{f(x)}{x}=f'(0). A limit of 0 therefore means f'(0)=0, and any nonzero limit is read directly as the first derivative at 0.

When the denominator is x^2, a Taylor expansion near 0 is used: f(x)=f(0)+f'(0)x+\frac{f''(0)}{2}x^2+\cdots. If f(0)=f'(0)=0, then \frac{f(x)}{x^2}\to\frac{f''(0)}{2}, so the limit equals half the second derivative at 0.

If \frac{u(x)}{v(x)} has a finite limit while v(x)\to 0, then u(x)\to 0 as well. The numerator must vanish at the point, which fixes the value of a differentiable function there. Continuity then gives the limit of the function as that same value.

When a relation links f(x) and f(1/x), replacing x by 1/x produces a second equation. Solving the two linear equations for f(x) gives an explicit formula, which can be differentiated and evaluated at the required point.

Limits of exponential expressions

A quotient of exponential expressions that gives \frac{0}{0} at a point is evaluated by L'Hôpital's rule, which replaces the quotient by the ratio of derivatives. For example, with \frac{a^x-x^a}{x^a-a^a} at x=a, one uses \frac{d}{dx}a^x=a^x\ln a and \frac{d}{dx}x^a=a\,x^{a-1}. Evaluating at x=a gives the limit \ln a-1.

For a ratio of exponentials with different bases, the larger base dominates as the exponent grows. If 0<b<a then (b/a)^n\to 0 as n\to\infty. Dividing numerator and denominator by a^n removes the dominant term, and the limit becomes \frac{1+0}{1-0}.

An indeterminate form 1^{\infty} is handled by writing L=\exp\left(\lim \frac{\ln f(x)}{g(x)}\right). When f(x)\to 1, \ln f(x) behaves like f(x)-1, so the exponent reduces to \lim\frac{f(x)-1}{g(x)}. L'Hôpital's rule then gives \frac{f'(1)}{g'(1)}. The derivatives of a^{x-1} and b^{x-1} each contribute a factor \ln a or \ln b, and their sum gives \ln(ab), so the exponent becomes \frac{\ln(ab)}{2}.

Finally, \exp\left(\frac{\ln(ab)}{2}\right)=\sqrt{ab}, which is the value of the 1^{\infty} limit.

Differentiability of absolute value functions

A function f is differentiable at a point a only when the left-hand derivative and the right-hand derivative at a both exist and are equal. Continuity at a is necessary but never sufficient. The modulus |x| is continuous everywhere, yet its graph has a sharp corner at x=0, where \lim_{h\to0^+}\frac{|h|}{h}=1 and \lim_{h\to0^-}\frac{|h|}{h}=-1. Therefore |x| is not differentiable at 0.

For a composite |g(x)|, the rule is direct. Where g(x)>0 the derivative is g'(x), and where g(x)<0 it is -g'(x). At points where g(x)=0, the one-sided derivatives must be checked directly. For example, on (0,1) we have \ln x<0, so |\ln x|=-\ln x and its derivative is -\frac{1}{x}.

The same one-sided test applies to any smooth function of |x|. For f(x)=e^{|x|}, the right-hand derivative at 0 is 1 while the left-hand derivative is -1, so f'(0) does not exist even though f is continuous there.

Piecewise functions built from straight segments follow the same logic. Continuity at a junction is checked by comparing the two formulas at that point, and differentiability by comparing the slopes of the two pieces. If the slopes differ, the graph has a corner and the derivative is undefined there, however smooth each piece is.

A common error is to assume that continuity at a point implies differentiability there. For the same function |x|-1, the statement 'it is continuous at x=1' can be true while 'it is differentiable at x=0' is false, so each statement must be tested separately.

Logarithmic differentiation

Logarithmic differentiation is used when a function is a product of many factors, or when both the base and the exponent depend on x. Take the natural logarithm of both sides, differentiate, and multiply back by y, using \frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}\ln|y|. For a product y=u_1u_2\cdots u_n, this gives \frac{1}{y}\frac{dy}{dx}=\sum_{i}\frac{u_i'}{u_i}, which avoids a long product-rule expansion.