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MP DTE Mathematics — practice by topic

Browse 1,043 questions across 156 Mathematics topics. Choose what you want to revise, then open a question to check its answer and worked solution.

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Evaluate $\displaystyle\lim_{x \to 0} \dfrac{\sin 3x}{x}$.Read the answer and worked solution →Evaluate $\displaystyle\lim_{x \to 2} \dfrac{x^2 - 4}{x - 2}$.Read the answer and worked solution →Evaluate $\displaystyle\lim_{x \to 0} \dfrac{1 - \cos 2x}{x^2}$.Read the answer and worked solution →Evaluate $\displaystyle\lim_{x \to \infty} \left(1 + \dfrac{2}{x}\right)^x$.Read the answer and worked solution →

Worked practice questions

Evaluate $\displaystyle\lim_{x \to 0} \dfrac{\sin 3x}{x}$.

  1. $1$
  2. $3$
  3. $\infty$
  4. $0$
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Correct answer: B. $3$

Using the standard limit $\lim_{u\to 0} \dfrac{\sin u}{u} = 1$, write $\dfrac{\sin 3x}{x} = 3 \cdot \dfrac{\sin 3x}{3x}$. As $x \to 0$, the second factor $\to 1$, giving $3$.

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Evaluate $\displaystyle\lim_{x \to 2} \dfrac{x^2 - 4}{x - 2}$.

  1. The limit does not exist
  2. $4$
  3. $2$
  4. $0$
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Correct answer: B. $4$

Factor the numerator: $x^2 - 4 = (x-2)(x+2)$. Cancelling, $\dfrac{(x-2)(x+2)}{x-2} = x + 2$ for $x \neq 2$. Limit = $4$. Option D is the trap if you stop at the $0/0$ form without factoring.

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Evaluate $\displaystyle\lim_{x \to 0} \dfrac{1 - \cos 2x}{x^2}$.

  1. $2$
  2. $4$
  3. $0$
  4. $1$
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Correct answer: A. $2$

Use $1 - \cos 2x = 2\sin^2 x$. Then $\dfrac{2\sin^2 x}{x^2} = 2\left(\dfrac{\sin x}{x}\right)^2 \to 2 \cdot 1 = 2$ as $x \to 0$. Alternatively, expand $\cos 2x \approx 1 - 2x^2$ to leading order, giving $\dfrac{2x^2}{x^2} = 2$.

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