IBDP Maths AA: Topic SL 5.1: concept of a limit: IB style Questions SL Paper 2

Question

[Maximum mark: 7]
Part of the graph of \(f(x)=\frac{e^{x-2}-x+1}{(x-2)^{2}}\) is shown below:

(a) Write down, correct to 4 s.f. , the value of f (0)
(b) Explain why \(f (2)\) is not defined.
(c) Complete the values of \(f\) , correct to 4 s.f. on the tables below:

(d) Deduce the value of the limit:       \(\lim_{x\to 2}\frac{e^{x-2}-x+1}{(x-2)^{2}}\)

Answer/Explanation

Ans.

(a) f (1) = 0.2838

(b) because the denominator is 0.

(c)

(d)  \(\lim_{x \to 2}\frac{e^{x-2}-x+1}{(x-2)^{2}}=0.5\)

Question

[Maximum mark: 7]
(a) By observing the graph of the function \(f(x)=\frac{x+3}{x-2}\)on your GDC, or otherwise
find the values of the following limits

(i)   \(\lim_{x \to 3}\frac{x+3}{x-2}\)           (ii)  \(\lim_{x \to +\infty }\frac{x+3}{x-2}\)            (iii) \(\lim_{x \to -\infty }\frac{x+3}{x-2}\)

(b) Investigate whether each of the following side limits is +∞ or -∞:

(i) \(\lim_{x \to 2^{+}}\frac{x+3}{x-2}\)                                 (ii) \(\lim_{x \to 2^{-}}\frac{x+3}{x-2}\)

 

(iii) \(\lim_{x \to 2^{+}}\frac{x-3}{x-2}\)                                    (iv)  \(\lim_{x \to 2^{-}}\frac{x-3}{x-2}\)

Answer/Explanation

Ans.

(a)    (i)   \(\lim_{x \to 3}\frac{x+3}{x-2}=6\)          (ii)  \(\lim_{x \to +\infty }\frac{x+3}{x-2}=1\)           (iii) \(\lim_{x \to -\infty }\frac{x+3}{x-2}=1\)

(b)   (i)  \(\lim_{x \to 2^{+}}\frac{x+3}{x-2}=+\infty\)                   (ii) \(\lim_{x \to 2^{-}}\frac{x+3}{x-2}=-\infty\)

(iii)  \(\lim_{x \to 2^{+}}\frac{x-3}{x-2}=-\infty\)                     (iv)  \(\lim_{x \to 2^{-}}\frac{x-3}{x-2}=+\infty\)

Question

[Maximum mark: 8]
The gradient of the curve \(f(x)=x^{2}\) at point x = a is defined by the limit

\(m_{a}=\lim_{h\to0}\frac{(a+h)^{2}-a^{2}}{h}\)

For example, the gradient at x = 1 is \(m_{1}=2\) since  \(\lim_{h\to0}\frac{(1+h)^{2}-1^{1}}{h}=2\).

(a) By using the graph mode on your GDC, find

(i) \(\lim_{h\to 0}\frac{(2+h)^{2}-2^{2}}{h}\) and hence the gradient \(m_{2}\).

(ii) \(\lim_{h\to 0}\frac{(3+h)^{2}-3^{2}}{h}\) and hence the gradient \(m_{3}\).

(b) Find the gradient of the curve
(i)   at \(x\) = 5.               (ii)\(x\) = – 5

(c) Deduce the value of the gradient \(m_{a}\) in terms of \(a\) in general.

Answer/Explanation

Ans.

(a)    (i) \(\lim_{h\to 0}\frac{(2+h)^{2}-2^{2}}{h}= 4\),        \(m_{2}= 4\)

(ii) \(\lim_{h\to 0}\frac{(3+h)^{2}-3^{2}}{h}= 6\),          \(m_{3}= 6\)

(b)    (i) \(m_{5}= 10\),                             (ii) \(m_{-5}= – 10\)

(c)  \(m_{a} = 2a\)

 

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