# Mathematics Questions and Answers – Limits

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This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Limits”.

1. What is the value of $$\lim\limits_{y \rightarrow 2} \frac{y^2-4}{y-2}$$?
a) 2
b) 4
c) 1
d) 0

Explanation: y2 – 4 = (y – 2)(y + 2)
Therefore the fraction becomes, (y + 2)
As y tends to 2, the fraction becomes 4

2. What is the value of $$\lim\limits_{y \rightarrow \infty} \frac{2}{y}$$?
a) 0
b) 1
c) 2
d) Infinity

Explanation: Any number divided by infinity gives us 0.
Here, since the number 2 is divided by y, as y approaches infinity, we get 0

3. What is the value of $$\lim\limits_{x \rightarrow 4} \frac{x^2-2x-8}{x-4}$$?
a) 0
b) 2
c) 8
d) 6

Explanation: The denominator becomes 0, as x approaches 4.
$$\lim\limits_{x \rightarrow 4} \frac{x^2-2x-8}{x-4}$$ Here, if we factorize the numerator we get:
$$\lim\limits_{x \rightarrow 4} \frac{(x – 4)(x + 2)}{x – 4}$$
We can now cancel out (x – 4) from both the numerator and denominator.
We get, $$\lim\limits_{x \rightarrow 4}$$(x + 2) = 6
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4. What is the value of $$\lim\limits_{x \rightarrow 3}\frac{x^2-9}{x–3}$$?
a) 0
b) 3
c) Infinity
d) 6

Explanation: When x tends to 3, both the numerator and the denominator become 0 and it becomes of the form, $$\frac{0}{0}$$.
Therefore, we use L’Hospital’s rule, which states the we differentiate the numerator and the denominator, until a definite answer is reached.
On differentiating once we get,
$$\lim\limits_{x \rightarrow 3}\frac{2x}{1}$$
Since, this not an indeterminate form now, we can substitute the value of x.
= 2 x 3
= 6

5. What is the value of $$\lim\limits_{x \rightarrow \infty}\frac{x^2-9}{x^2–3x+2}$$?
a) 1
b) 2
c) 0
d) Limit does not exist

Explanation:
Since it is of the form $$\frac{\infty}{\infty}$$, we use L’Hospital’s rule and differentiate the numerator and denominator
L = $$\lim\limits_{x \rightarrow \infty}\frac{x^2-9}{x^2–3x+2}$$
On differentiating once, we get $$\lim\limits_{x \rightarrow \infty}\frac{2x}{2x}$$
Which is equal to, $$\lim\limits_{x \rightarrow \infty}$$ ⁡1 = 1.
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6. Which of the following limits does not yield 1?
a) $$\lim\limits_{x \rightarrow 0}$$⁡ 1
b) $$\lim\limits_{x \rightarrow \infty}$$x-2 + x-1 + 1
c) $$\lim\limits_{x \rightarrow \infty}\frac{1}{e^x}$$ + 1
d) $$\lim\limits_{x \rightarrow \infty}\frac{x^3+x^2+32x+1}{x^2–3x+2}$$

Explanation: $$\lim\limits_{x \rightarrow 0}$$⁡ 1 = 1 (Since no x term is present)
When the denominator is infinity, the value of the fraction is 0, provided the numerator is not infinity.
$$\lim\limits_{x \rightarrow \infty}$$x-2 + x-1 + 1 = 0 + 0 + 1 = 1
$$\lim\limits_{x \rightarrow \infty}\frac{1}{e^x}$$ + 1 = 1 ( e-∞ = 0)
$$\lim\limits_{x \rightarrow \infty}\frac{x^3+x^2+32x+1}{x^2–3x+2}$$ (Use L’Hospital’s rule and differentiate the numerator and denominator until a rational form is obtained)
$$\lim\limits_{x \rightarrow \infty}\frac{x^3+x^2+32x+1}{x^2–3x+2}$$ = $$\lim\limits_{x \rightarrow \infty}\frac{3x^2+2x+32}{2x–3}$$ = $$\lim\limits_{x \rightarrow \infty}\frac{3x}{2}$$ = ∞

7. What is the value of $$\lim\limits_{y \rightarrow 4}$$ f(y)? It is given that f(y) = y2 + 6y (y ≥ 2) and f(y) = 0 (y < 2).
a) 40
b) 16
c) 0
d) 30

Explanation: $$\lim\limits_{y \rightarrow 4}$$f(y) = y2 + 6y
f(4) = 42 + 6(4)
f(4) = 16 + 24
f(4) = 40

8. What is the value of the limit f(x) = $$\frac{x^2+\sqrt {2x}}{x^2-4x}$$ if x approaches infinity?
a) 0
b) 2
c) 1/2
d) 4

Explanation: This is of the form $$\frac{\infty}{\infty}$$, therefore we use L’Hospital’s rule and differentiate the numerator and denominator.
= $$\lim\limits_{x \rightarrow \infty}\frac{2x+\sqrt{2/x}}{2x–4}$$
= $$\lim\limits_{x \rightarrow \infty}⁡$$√2 x-3/2
= 0

9. What is the value of the $$\lim\limits_{x \rightarrow 5}⁡\frac{32x+1}{x^2–5x}$$?
a) 6.2
b) 6.4
c) 6.3
d) 6.1

Explanation: Use L’Hospital’s Rule, and differentiate the numerator and denominator.
$$\lim\limits_{x \rightarrow 5}\frac{⁡32}{2x–5}$$
= $$\frac{32}{5}$$
= 6.4

10. What is the value of the limit $$\lim\limits_{x \rightarrow 4}\frac{x^2-4-3x}{x-3}$$?
a) 0
b) 4
c) 1
d) Limit does not exist

Explanation: $$\lim\limits_{x \rightarrow 4}\frac{x^2-4-3x}{x-3}$$
= $$\frac{4^2-4-3(4)}{4-3}$$
= $$\frac{0}{1}$$
= 0.

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