"Limits, Continuity and Differentiability" — Practice

Question 1

Evaluate the limit of (x cubed minus 8) divided by (x squared minus 4) as x approaches 2.

Accepted Answer

Substituting gives zero over zero, which signals a rewrite rather than an answer. Both parts vanish at x equal to 2, so both contain a factor of x minus 2. The numerator is a difference of cubes and factors as (x minus 2) times (x squared plus 2x plus 4). The denominator is a difference of squares and factors as (x minus 2) times (x plus 2). Cancelling the shared factor is legitimate because x approaches 2 without equalling it. The expression becomes (x squared plus 2x plus 4) over (x plus 2), and substituting now gives 12 over 4, which is 3. Note that the function itself remains undefined at x equal to 2; the limit describes the approach.

Question 2

Evaluate the limit of (1 minus cos 2x) divided by x squared as x approaches 0.

Accepted Answer

Substitution gives zero over zero and no factor is available to cancel, so convert to a standard limit. Use the identity that 1 minus cos 2x equals 2 sin squared x. The expression becomes 2 sin squared x over x squared, which is 2 times the square of (sin x over x). As x approaches zero the bracket tends to 1, so the whole expression tends to 2. Checking against the general standard limit, since (1 minus cos t) over t squared tends to one half, putting t equal to 2x gives (1 minus cos 2x) over 4x squared tending to one half, so the original tends to 2. Both routes assume radian measure.

Question 3

Evaluate the limit of (root of (x plus 4) minus 2) divided by x as x approaches 0.

Accepted Answer

Substituting gives zero over zero. Because the vanishing difference involves a square root, rationalise by multiplying numerator and denominator by the conjugate, which is root of (x plus 4) plus 2. The numerator becomes (x plus 4) minus 4, which is simply x. The expression is therefore x divided by x times the quantity root of (x plus 4) plus 2, and the x cancels. What remains is 1 over the quantity root of (x plus 4) plus 2. Substituting x equal to zero gives 1 over (2 plus 2), which is one quarter.

Question 4

Evaluate the limit of (3x squared minus 2x plus 1) divided by (5x squared plus 7) as x grows without bound, and state the general rule.

Accepted Answer

Divide both the numerator and the denominator by x squared, the highest power present. The numerator becomes 3 minus 2 over x plus 1 over x squared, and the denominator becomes 5 plus 7 over x squared. Every term carrying a negative power of x vanishes as x grows without bound, leaving 3 over 5. The general rule for a ratio of polynomials follows from the same division: if the numerator has the smaller degree the limit is zero, if the degrees are equal the limit is the ratio of the leading coefficients, and if the numerator has the larger degree there is no finite limit.

Question 5

Show that f(x) equal to the modulus of x minus 2 is continuous at x equal to 2 but not differentiable there.

Accepted Answer

For continuity, both one-sided limits approach zero as x approaches 2, and the function value there is the modulus of zero, which is zero. All three requirements hold, so the function is continuous at that point. For differentiability, compute the one-sided derivatives. Approaching from the left, x minus 2 is negative so the function equals minus (x minus 2) and its slope is minus one. Approaching from the right, x minus 2 is positive so the function equals x minus 2 and its slope is plus one. The two sides agree about the value but disagree about the slope, so the derivative does not exist at x equal to 2. This is the implication chain in one example, with the graph showing a corner at that point.

Question 6

Find a and b so that f(x) equal to 3x plus a for x at most 1, and bx squared plus 5 for x greater than 1, is differentiable at x equal to 1.

Accepted Answer

Check continuity first, since differentiability cannot hold without it. The left piece gives the value 3 plus a at x equal to 1, and the right piece approaches b plus 5, so continuity requires 3 plus a to equal b plus 5. Now compare slopes. The left piece has derivative 3 everywhere. The right piece has derivative 2bx, which is 2b at x equal to 1. Differentiability therefore requires 3 to equal 2b, so b is three halves. Substituting into the continuity equation gives 3 plus a equal to three halves plus 5, so a is seven halves. Checking, both pieces give the value 6.5 at x equal to 1 and both derivatives equal 3. Solving the slope equation first gave b immediately, avoiding simultaneous equations.

Question 7

Prove that a function differentiable at a point must be continuous there, and explain why the converse fails.

Accepted Answer

Suppose the derivative at a exists, meaning the limit of [f(a plus h) minus f(a)] over h as h approaches zero is some finite number L. Write f(a plus h) minus f(a) as the quotient multiplied by h. As h approaches zero, the quotient approaches L and h approaches zero, so by the product rule for limits the whole expression approaches L times zero, which is zero. That says f(a plus h) approaches f(a), which is exactly the statement that f is continuous at a. The converse fails because continuity requires only that the two sides agree about the value, while differentiability additionally requires them to agree about the slope. At a corner, such as the modulus function at the origin, the values agree but the slopes are minus one and plus one, so the stronger condition fails while the weaker one holds.

Question 8

Is f(x) equal to the cube root of x differentiable at x equal to 0? Justify your answer.

Accepted Answer

The function is continuous at zero, since the cube root of x approaches zero from both sides and the function takes the value zero there. For differentiability, apply the definition: the difference quotient is the cube root of h divided by h, which equals h to the power minus two thirds. As h approaches zero this grows without bound rather than settling on a finite value, so the limit does not exist and the function is not differentiable at zero. Geometrically the graph has a vertical tangent at the origin. This is the second failure mode: unlike a corner, where two finite slopes disagree, here the slope is unbounded, yet the function remains perfectly continuous.

Question 9

Differentiate y equal to x to the power sin x, for x greater than zero.

Accepted Answer

Neither the power rule nor the exponential rule applies, because the variable appears in both the base and the exponent, so use logarithmic differentiation. Taking logarithms of both sides gives log y equal to sin x times log x, which turns the exponent into a multiplier. Differentiating both sides with respect to x, the left gives one over y times dy by dx by the chain rule, and the right needs the product rule, giving cos x times log x plus sin x over x. Multiplying through by y and substituting the original expression back gives dy by dx equal to x to the power sin x, times the quantity cos x log x plus sin x over x. Substituting y back is a required step; leaving the answer in terms of y is incomplete.

Question 10

Find dy by dx for the curve x squared plus xy plus y squared equal to 7.

Accepted Answer

Differentiate every term with respect to x, treating y as a function of x. The first term gives 2x. The second term is a product of x and y, so the product rule gives y plus x times dy by dx. The third term needs the chain rule, giving 2y times dy by dx. The right side is constant so it gives zero. Collecting, 2x plus y plus x dy by dx plus 2y dy by dx equals zero. Grouping the derivative terms gives dy by dx times (x plus 2y) equal to minus (2x plus y). So dy by dx equals minus (2x plus y) over (x plus 2y). The answer containing both variables is expected, since the curve is not the graph of a single function of x and the slope genuinely depends on the point.

Question 11

Differentiate y equal to the inverse tangent of (2x divided by (1 minus x squared)) with respect to x, for the modulus of x less than 1.

Accepted Answer

Applying the chain rule and quotient rule directly is possible but slow. Recognise the argument as the double-angle form for tangent: if x equals tan theta then 2x over (1 minus x squared) equals tan 2 theta. So y equals the inverse tangent of tan 2 theta, which for the modulus of x less than 1 gives y equal to 2 theta, that is 2 times the inverse tangent of x. Differentiating now is immediate: dy by dx equals 2 divided by (1 plus x squared). The restriction on x matters, because outside that range the inverse tangent returns a value differing by pi and the simplification would be wrong, although the derivative would still be the same.

Question 12

If y equals sin(3x) times e to the power 2x, find dy by dx, and then find the second derivative of y equals x to the power 4 minus 3x squared.

Accepted Answer

For the first, use the product rule with u equal to sin 3x and v equal to e to the power 2x. The derivative of sin 3x is 3 cos 3x by the chain rule, and the derivative of e to the power 2x is 2 e to the power 2x, again by the chain rule. So dy by dx equals 3 cos 3x times e to the power 2x plus sin 3x times 2 e to the power 2x, which factors as e to the power 2x times the quantity 3 cos 3x plus 2 sin 3x. For the second, differentiate x to the fourth minus 3x squared to get 4x cubed minus 6x, then differentiate again to get 12x squared minus 6. The syllabus goes no further than second derivatives.

Practice — "Limits, Continuity and Differentiability"

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