Continuity and Differentiability

• Mistake 1: Checking only \(\lim_{x\rightarrow c^-}f(x)=\lim_{x\rightarrow c^+}f(x)\). This proves existence of the two-sided limit, not continuity.
• Mistake 2: Forgetting to calculate \(f(c)\).
• Mistake 3: Substituting \(x=c\) directly into an expression containing \(0/0\) and declaring the function discontinuous without evaluating the limit.
• Mistake 4: Cancelling a factor and forgetting that the original function may still be undefined at the cancelled point.
• Mistake 5: Using the wrong branch of a piecewise function to calculate \(f(c)\).
• Mistake 6: Assuming that every function that looks smooth in a rough graph is mathematically continuous.

A frequent error is to write

Students sometimes calculate the limit at only one particular value of \(x\) and conclude that the function is continuous everywhere. To prove continuity on \(\mathbb{R}\) directly from the definition, \(c\) must be treated as an arbitrary real number.

Here, taking \(c\in\mathbb{R}\) arbitrarily gives

which establishes continuity at every point.

• Mistake 1: Writing \(f(0)=1/0\). This value is undefined.
• Mistake 2: Claiming that \(1/x\) is continuous everywhere on \(\mathbb{R}\).
• Mistake 3: Forgetting to state the domain.
• Mistake 4: Treating \(x=0\) as an ordinary point of continuity even though it is not in the domain.

• Calculating \(f(1)\) from \(x-2\). This is incorrect because the condition for the second branch is \(x>1\).
• Checking only \(f(1)=3\) and the left-hand limit. The right-hand limit must also be checked.
• Saying that the function is discontinuous everywhere because it is piecewise defined. Each individual branch is continuous on its own interval.
• Forgetting that unequal LHL and RHL immediately imply that the two-sided limit does not exist.
• Calling this a removable discontinuity. Since the LHL and RHL are unequal, the discontinuity is a jump discontinuity.

• Writing \(f(0)=2\) for the original function. The original function does not define \(f(0)\).
• Calling the original function discontinuous at \(0\) without mentioning that \(0\notin\operatorname{Domain}(f)\).
• Assuming that the existence of the limit automatically means continuity.
• Forgetting to calculate both one-sided limits at the point where the formula changes.
• Calling the discontinuity a jump discontinuity. Here the LHL and RHL are equal, so the missing point is removable.

• Using \(x^2\) to calculate \(f(0)\). The branch \(x<0\) excludes \(0\).
• Checking only the two one-sided limits and forgetting the actual value \(f(0)\).
• Assuming a change in formula necessarily produces discontinuity.
• Saying that continuity at \(0\) automatically implies differentiability at \(0\).
• Writing \(x>0\) when the given condition is \(x\geq0\). The equality sign determines which branch defines \(f(0)\).

• Assuming \([x]\) is discontinuous at every real number because its graph is made of steps. It is continuous inside every step.
• Writing \(\lim_{x\rightarrow n^-}[x]=n\). The correct value is \(n-1\).
• Confusing the greatest integer function with the nearest-integer function.
• Forgetting that negative numbers require special care. For example, \([-1.2]=-2\).
• Counting non-integers when a question asks for the discontinuities of \([x]\).

Do not write:

The correct statement is:

The distinction is important because the denominator may vanish at one or more real numbers.

• Writing \(\lim_{h\rightarrow0}\sin h=1\). The correct result is \(0\).
• Writing \(\lim_{h\rightarrow0}\cos h=0\). The correct result is \(1\).
• Forgetting to use the sine addition formula:
  \[ \sin(c+h)=\sin c\cos h+\cos c\sin h. \]
• Checking only \(f(c)\) and not comparing it with the limiting value.
• Claiming that every trigonometric function is continuous on all of \(\mathbb R\). Functions such as \(\tan x\), \(\sec x\), \(\cot x\) and \(\cosec x\) have points where they are undefined.

• Saying that \(f\circ g\) is continuous merely because \(f\) is continuous. Continuity of the inner function \(g\) at \(c\) is also required.
• Checking \(f\) at \(c\) instead of at \(g(c)\). The correct point for the outer function is \(g(c)\).
• Confusing \(f\circ g\) with \(g\circ f\).
• Ignoring domain restrictions of the outer function.
• Splitting a complicated composite function into cases unnecessarily when known continuity theorems already establish the result.

• Writing \(f'(x)=\lim_{h\rightarrow0}f(x+h)-f(x)/h\) without proper brackets. The correct expression is
  \[ f'(x)=\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}. \]
• Forgetting that the derivative exists only when the required limit exists as a finite value.
• Using the product rule incorrectly as \((uv)'=u'v'\).
• Forgetting the square on the denominator in the quotient rule.
• Assuming that a continuous function must be differentiable.
• Ignoring points where a trigonometric function is undefined.
• Using degree measure instead of radian measure in standard trigonometric derivative formulas. The standard calculus formulas are based on radians.

• Writing
  \[ \lim_{x\rightarrow c} \frac{f(x)-f(c)}{c} \]
  instead of the correct denominator \(x-c\).
• Forgetting that the derivative must exist as a finite number.
• Claiming that continuity implies differentiability. This is false.
• Checking only the value of \(f(c)\) when testing differentiability.
• Forgetting that a function such as \(|x|\) can be continuous at a point but fail to have a derivative there.
• Using \(f'(x)\) where the proof requires the specific value \(f'(c)\).

• Forgetting to multiply by the derivative of the inner function.
• Writing
  \[ \frac{d}{dx}(2x+1)^3=3(2x+1)^2 \]
  instead of the correct
  \[ 6(2x+1)^2. \]
• Differentiating only the outermost function in a nested expression.
• Confusing
  \[ f(g(x)) \]
  with
  \[ f(x)g(x). \]
• Expanding a complicated composite function unnecessarily.
• Forgetting the negative sign in
  \[ (\cos x)'=-\sin x. \]

Do not write

The correct result is

Also remember that \(\pi\) is a constant, so

A frequent error is to write

This is incorrect because \(y\) is not the independent variable. Since \(y\) depends on \(x\), the Chain Rule requires

Similarly,

Verification by Differentiating the Explicit Relation

The equation

generally cannot be conveniently solved for \(y\) using elementary functions. Therefore, implicit differentiation is much more efficient than attempting to isolate \(y\).

This illustrates an important advantage of implicit differentiation: the dependent variable need not be explicitly expressed in terms of the independent variable.

• Confusing \(\sin^{-1}x\) with \(\frac{1}{\sin x}\).
• Forgetting the negative sign in
  \[ \frac{d}{dx}(\cos^{-1}x) = -\frac{1}{\sqrt{1-x^2}}. \]
• Forgetting the derivative of the inner function while applying the Chain Rule.
• Writing
  \[ \frac{d}{dx}[\sin^{-1}(u)] = \frac{1}{\sqrt{1-u^2}} \]
  instead of
  \[ \frac{u'}{\sqrt{1-u^2}}. \]
• Ignoring the domain restrictions of inverse trigonometric functions.
• Incorrectly replacing \(\sqrt{x^2}\) by \(x\). In general,
  \[ \boxed{\sqrt{x^2}=|x|}. \]

• Writing
  \[ \frac{d}{dx}(a^x)=a^x \]
  for an arbitrary base \(a\). The correct result is
  \[ \boxed{\frac{d}{dx}(a^x)=a^x\ln a}. \]
• Forgetting the Chain Rule:
  \[ \frac{d}{dx}\ln(x^2+1) \neq\frac{1}{x^2+1}. \]
  The correct derivative is
  \[ \boxed{\frac{2x}{x^2+1}}. \]
• Assuming \(\log(x+y)=\log x+\log y\). This is false. The product rule applies to multiplication:
  \[ \boxed{\log(xy)=\log x+\log y}. \]
• Assuming \(\log(x-y)=\log x-\log y\). This is also false.
• Ignoring the restriction \(x>0\) for \(\ln x\) in real calculus.
• Confusing inverse logarithmic and exponential relationships.

• Do not write \(\dfrac{d}{dx}(e^{\cos x})=e^{\cos x}\). The factor \(-\sin x\) must also be included.
• Do not forget that \(\dfrac{d}{dx}(-x)=-1\).
• For \(\ln u\), use \(\dfrac{u'}u\), not merely \(\dfrac1u\).
• For \(\cos^{-1}u\), the derivative is negative:
  \[ -\frac{u'}{\sqrt{1-u^2}}. \]
• When an inverse trigonometric function contains another function as its argument, always apply the Chain Rule.

• Applying the ordinary power rule directly to \(x^x\). The exponent is not constant, so \(d(x^n)/dx=nx^{n-1}\) cannot be used with \(n=x\).
• Writing
  \[ \frac{d}{dx}\ln(u)=\frac1u \]
  instead of
  \[ \frac{d}{dx}\ln(u)=\frac{u'}u. \]
• Forgetting the Product Rule in
  \[ v(x)\ln(u(x)). \]
• Forgetting to multiply by \(y\) after obtaining
  \[ \frac{y'}y. \]
• Confusing \(\ln(u^v)\) with \(v\ln u\) without considering the domain. For real-valued logarithmic differentiation, the logarithm must be applied where the expression is defined and positive.

• Forgetting the factor \(\frac12\) arising from the square root.
• Writing \(\ln(AB)=\ln A\,\ln B\). The correct identity is
  \[ \ln(AB)=\ln A+\ln B. \]
• Writing
  \[ \ln\left(\frac AB\right)=\frac{\ln A}{\ln B}. \]
  The correct identity is
  \[ \ln\left(\frac AB\right)=\ln A-\ln B. \]
• Forgetting the Chain Rule while differentiating \(\ln(x^2+4)\) or \(\ln(3x^2+4x+5)\).
• Forgetting that
  \[ \frac{d}{dx}\ln[u(x)]=\frac{u'(x)}{u(x)}. \]

• Incorrect:
  \[ \frac{d}{dx}(x^{\sin x})=\sin x\,x^{\sin x-1}. \]
  This incorrectly treats \(\sin x\) as a constant.
• Incorrect:
  \[ \frac{d}{dx}(\sin x\ln x)=\cos x\ln x. \]
  The second Product Rule term has been omitted.
• Correct:
  \[ \frac{d}{dx}(\sin x\ln x) = \cos x\ln x+\frac{\sin x}{x}. \]
• Do not confuse \(\log x\) with \(x\log x\). The derivative of \(\log x\), when \(\log\) denotes the natural logarithm, is \(1/x\).

• Writing
  \[ \frac{d}{dx}(\cos^{-1}x)=-\sin^{-1}x. \]
  This is incorrect.
• Writing
  \[ \frac{d}{dx}(\sin^{-1}x)=\cos^{-1}x. \]
  This is also incorrect.
• Forgetting that \(\sin^{-1}x\) means the inverse sine function, not \(\dfrac{1}{\sin x}\).
• Using \(\dfrac{dy}{d\theta}/\dfrac{dx}{d\theta}\) without checking that \(\dfrac{dx}{d\theta}\neq0\).

• Using
  \[ \frac{dy}{dx}=\frac{dx/dt}{dy/dt}. \]
  The numerator must be \(dy/dt\).
• Forgetting that \(a\) is a constant.
• Failing to check the condition
  \[ \frac{dx}{dt}\neq0. \]
• At \(t=0\), substituting directly into \(\dfrac{1}{t}\) and concluding that the derivative does not exist without examining the tangent geometrically.