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) \(m_{a} = 2a\)