Limits and Derivatives MCQs

Substitute the approaching value into the linear expression: 2·3 + 1.

Compute the multiplication: 2·3 = 6.

Add the constant term: 6 + 1 = 7.

Hence the limit equals 7.

Use the standard limit \(\displaystyle \lim_{x\to0}\frac{\sin kx}{kx}=1\).

Rewrite \(\frac{\sin(3x)}{\sin(5x)}=\frac{3x}{5x}\cdot\frac{\sin(3x)}{3x}\cdot\frac{5x}{\sin(5x)}\).

As \(x\to0\), \(\frac{\sin(3x)}{3x}\to1\) and \(\frac{5x}{\sin(5x)}\to1\).

Thus the limit equals \(\frac{3}{5}\).

Use the identity 1‑cos x = 2 sin²(x/2).

Then (1‑cos x)/x = 2 sin²(x/2)/x = (sin(x/2)/(x/2))·sin(x/2).

As x→0, sin(x/2)/(x/2) → 1 and sin(x/2) → 0, so the product tends to 0.

Hence the limit equals 0.

Apply power rule: d/dx[x^n]=n x^{n-1}.

Derivative of 2x^3 is 2·3 x^{2}=6x^2.

Derivative of -3x is -3·1 x^{0}= -3.

Add the two results: 6x^2 - 3.

Answer: 6x^2 - 3

Use product rule: d/dx[(x^2+1)(x+1)] = (x^2+1)·d/dx(x+1) + (x+1)·d/dx(x^2+1).

Derivative of (x+1) is 1, derivative of (x^2+1) is 2x.

So expression becomes (x^2+1)·1 + (x+1)·2x = x^2+1 + 2x^2+2x = 3x^2+2x+1.

Hence the correct answer is 3x^2 + 2x + 1.

Use the standard limit \(\displaystyle \lim_{u\to0}\frac{\sin u}{u}=1\).

Let \(u=5x\); then as \(x\to0\), \(u\to0\) and \(\frac{\sin(5x)}{x}= \frac{\sin u}{u}\cdot5\).

Since \(\frac{\sin u}{u}\to1\), the whole expression tends to \(5\).

Hence the limit equals 5.

Use the standard limit limₓ→0 sin x / x = 1.

Rewrite sin x/(2x) = (1/2)·(sin x / x).

Since sin x / x → 1 as x→0, the whole expression → (1/2)·1 = 1/2.

Hence the limit equals 1/2.

Factor numerator: x^2−1 = (x−1)(x+1).

Cancel the common factor (x−1) with denominator, leaving x+1 for x≠1.

Substitute x=1 into the simplified expression: 1+1 = 2.

Hence the limit equals 2.

Use the limit lim_{x→0} sin kx / (kx) = 1, so sin kx ≈ kx for small x.

Rewrite sin(7x)/sin(4x) = (7x)/(4x)·[sin(7x)/(7x)]·[4x/sin(4x)].

Each bracket tends to 1 as x→0, leaving 7/4.

Thus the limit equals 7/4.

Use product rule: d/dx(sin x·cos x)= (cos x)(cos x)+(sin x)(-sin x).

Simplify: cos²x − sin²x.

Recall identity cos 2x = cos²x − sin²x.

Hence the derivative equals cos 2x.

Factor numerator: x^2‑4 = (x‑2)(x+2)

Cancel the common factor (x‑2) with denominator (x‑2) for x≠2, giving limit = x+2

Substitute x→2: 2+2 = 4

Hence the correct answer is 4

Factor the numerator: x²‑5x+6 = (x‑2)(x‑3).

Cancel the common factor (x‑2) with the denominator, giving the simplified expression x‑3 for x≠2.

Substitute x = 2 into x‑3 to get 2‑3 = -1.

Hence the limit equals -1.

Write the function as x⁻¹.

Differentiate using the power rule: d/dx [xⁿ] = n xⁿ⁻¹, with n = –1.

Thus d/dx [x⁻¹] = (–1) x^(–2) = –1/x².

The correct answer is –1/x².

Factor x from numerator: (x^2+5x)=x(x+5).

Cancel the common x with denominator x, giving limit of (x+5) as x→0.

Substitute x=0 into the simplified expression: 0+5 = 5.

Hence the limit equals 5.

Using the small‑angle approximation, sin x ≈ x when x is measured in radians.

Thus the ratio sin x / x approaches 1 as x→0.

Therefore the limit equals 1.

Apply the power rule: d/dx [x^n] = n·x^(n‑1).

Here n = 5, so differentiate x^5 → 5·x^(5‑1).

Simplify the exponent: 5·x^4.

Thus the derivative is 5x^4.

Write the function as x^{1/2}.

Differentiate using the power rule: d/dx x^{n}=n x^{n‑1}.

Here n=1/2, so derivative = (1/2) x^{1/2‑1} = (1/2) x^{-1/2}.

Rewrite x^{-1/2} as 1/√x, giving 1/(2√x).

Answer: 1/(2 sqrt(x))

cos x is continuous at x=0, so the limit equals its value at 0.

Substitute x=0: cos 0 = 1.

Hence limₓ→0 cos x = 1.

f'(x)=limₕ→0 [7(x+h)²−7x²]/h

Expand numerator: 7[x²+2xh+h²−x²]=7(2xh+h²)

Factor h: 7h(2x+h) /h =7(2x+h) → as h→0 gives 7·2x = 14x.

Hence the derivative is 14x.

Use the identity 1‑cos x = 2 sin²(x/2).

Rewrite the limit as 2·sin²(x/2)/x² = 2·[sin(x/2)/(x/2)]²·(1/4).

Since sin θ/θ → 1 as θ→0, the bracket tends to 1, giving 2·(1)·(1/4) = 1/2.

Hence the limit equals 1/2.