The midterm questions are similar to that of past exams. Please make sure you understand all of them. All examples in this page are selected from past exams.
\[lim_{x \to a} f(x) = f(a)\]
In other words, we do direct substitution if function is continuous at that point.
\[lim_{t \to 2}(6 - t^3)(t - 3) = (6 - 2^3)(2 - 3) = 2\]
Some techniques needed to simplify: taking common denominator, factor out polynomials, expanding out powers, multiplying conjugate forms…
\[\lim_{x \to 0} \frac{x}{\sqrt{x +3} - \sqrt{3}} = \lim_{x\to 0} \frac{x}{\sqrt{x +3} - \sqrt{3}} \cdot \frac{\sqrt{x +3} +\sqrt{3}}{\sqrt{x +3} +\sqrt{3}} = \lim_{x \to 0} \frac{\sqrt{x +3} +\sqrt{3}}{1} = 2 \sqrt{3}\]
\[\lim_{u \to 5} \frac{\sqrt{u+11} -4}{u -5} = \lim_{u\to 5} \frac{\sqrt{u+11} -4}{u -5} \cdot \frac{\sqrt{u+11} + 4}{\sqrt{u+11} + 4} = \lim_{u \to 5} \frac{1}{\sqrt{u+11} + 4} = \frac{1}{8}\]
\[\lim_{x \to -1^{+}}\frac{x^2-x -2}{x+1} = \lim_{x \to -1^{+}}\frac{(x+1)(x -2)}{x+1} = \lim_{x \to -1^{+}} (x -2) = -3\]
\[\lim_{x \to 3}\frac{x^2-2x -3}{x-3} =\lim_{x \to 3}\frac{(x-3)(x+1)}{x-3} = \lim_{x \to 3} (x +1) = 4\]
This happens when denominator becomes zero but numerator doesn’t. If one of \(\lim_{x \to a} f(x), \lim_{x \to a^{-}} f(x), \lim_{x \to a^{+}} f(x)\) is either \(\infty\), or \(-\infty\), \(x = a\) is called vertical asymptote.
\[\lim_{x \to 1^{-}}\frac{-2 \pi^x}{(x -1)^5}\]
Left-side limit means \(x < 1\), so \((x-1)\) is negative and so is \((x-1)^5\). The numerator \(-2 \pi^x\) stays negative, so the limit is \(\infty\).
\[f(x) = \frac{2 x^2 -2}{x^2 +x}\]
The domain of \(f(x)\) is when \(x^2 +x \neq 0\), i.e.,\(x \neq 0, -1\).
\[\lim_{x \to 0^{+}}\frac{2 x^2 -2}{x^2 +x} = \lim_{x \to 0^{+}}\frac{2 (x+1)(x-1)}{x(x+1)} = \lim_{x \to 0^{+}}\frac{2(x-1)}{x} = -\infty\]
\[\lim_{x \to 0^{-}}\frac{2 x^2 -2}{x^2 +x} = \lim_{x \to 0^{-}}\frac{2 (x+1)(x-1)}{x(x+1)} = \lim_{x \to 0^{-}}\frac{2(x-1)}{x} = \infty\]
So \(x = 0\) is a vertical asymptote.
\[\lim_{x \to -1}\frac{2 x^2 -2}{x^2 +x} = \lim_{x \to -1}\frac{2 (x+1)(x-1)}{x(x+1)} = \lim_{x \to -1}\frac{2(x-1)}{x} = 4\]
So \(x = -1\) is not a vertical asymptote.
\[f(x) = \frac{x^2 -2x}{x^2 -x-2}\]
The domain is when \(x^2 - x -2 \neq 0\), i.e., \(x \neq 2, -1\).
\[\lim_{x \to -1^{+}} \frac{x^2 -2x}{x^2 -x-2} = \lim_{x \to -1^{+}} \frac{x(x-2)}{(x-2)(x+1)} = \lim_{x \to -1^{+}}\frac{x}{x+1} = -\infty\]
\[\lim_{x \to -1^{-}}\frac{x^2 -2x}{x^2 -x-2} = \lim_{x \to -1^{-}}\frac{x(x-2)}{(x-2)(x+1)} = \lim_{x \to -1^{-}}\frac{x}{x+1} = \infty\]
So \(x = -1\) is a vertical asymptote.
\[\lim_{x \to 2}\frac{x^2 -2x}{x^2 -x-2} = \lim_{x \to 2}\frac{x(x-2)}{(x-2)(x+1)} = \lim_{x \to 2}\frac{x}{x+1} =\frac{2}{3}\]
So \(x = 2\) is not a vertical asymptote.
If \(\lim_{x \to \infty} f(x) = L\) or \(\lim_{x \to -\infty} f(x) = L\), \(y = L\) is called the horizontal asymptote. A useful technique is dividing both the numerator and denominator the highest power of \(x\) occurs in the denominator.
\[\lim_{x \to \infty}\frac{x^2 -2x}{x^2 -x-2} = \lim_{x \to \infty}\frac{1 -\frac{2x}{x^2}}{1 -\frac{x}{x^2}-\frac{2}{x^2}} = \frac{1}{1} = 1\]
Also \(\lim_{x \to -\infty}\frac{x^2 -2x}{x^2 -x-2} =1\). So \(y = 1\) is the horizontal asymptote.
\[\lim_{x \to \infty}\frac{2 x^2 -2}{x^2 +x} = \lim_{x \to \infty}\frac{2 -\frac{2}{x^2}}{1 +\frac{x}{x^2}} = \frac{2}{1} = 2\]
Also \(\lim_{x \to -\infty}\frac{2 x^2 -2}{x^2 +x} =2\). So \(y = 2\) is the horizontal asymptote.
This is useful when the question has \(\sin(\dot)\), or \(\cos(\cdot)\) as part of a function.
\[ \lim_{x \to \infty} \frac{1}{x} \sin(x) = 0\].
This is because \(-\frac{1}{x} \leq \frac{1}{x} \sin(x) \leq \frac{1}{x}\) and \(\lim_{x \to \infty} \frac{1}{x} = 0\)
Most functions we’ve seen are continuous in their domains.
\[\frac{x-1}{x+2}\]
The domain is when \(x \neq -2\). So function is continuous when \(x \neq -2\).
\[\log_{10}(x-1)\]
The domain is when \(x > 1\). So function is continuous when \(x > 1\).
\[\sqrt{x+2}\]
The domain is when \(x \geq -2\). So function is continuous when \(x \geq -2\).
To prove an equation \(f(x) = L\) has a solution or at least one root in the interval \((a, b)\). Need to show that \(f(x)\) is at least continuous on \([a, b]\), and value \(L\) is between \(f(a)\) and \(f(b)\).
\[f(x) = \sqrt{1+x} - \sqrt{2-x}\]
As function \(f(x)\) has domain as \(-1 \leq x \leq 2\), it is continuous on \([-1, 2]\). To show equation \(f(x) =1\) has a solution in \((-1, 2)\), check that \[f(-1) < 1 < f(2).\]
If \(a\) if cut-off point of the piecewise function \(f(x)\), it is continuous at \(a\) if \(\lim_{x \to a^{-}} f(x) = \lim_{x \to a^{+}} f(x) = f(a)\).
Let \[f(x) = \begin{cases} kx +2, \ \ x \leq 1 \\ x^2, \ \ x > 1 \end{cases}\]
For \(f(x)\) to be continuous at \(a = 1\), set \[ \lim_{x \to 1^{-}} kx+2 = \lim_{x \to 1^{+}} x^2\] so \(k+2 = 1\), \(k = -1\).
Know the statement of precise definition: for any positive number \(\epsilon\), there is another positive number \(\delta\), such that if \[0 < |x -a| < \delta,\] \[|f(x) - L| < \epsilon.\]
Prove \[\lim_{x \to 2} 3x-1 = 5\]
Let \(|(3x-1) -5| < \epsilon\), then \(3|x-2| < \epsilon\), \(|x-2| < \epsilon/3\). Therefore, for any \(\epsilon\), we found \(\delta = \epsilon/3\), if \[0 < |x -2| < \delta,\] \[|(3x-1) - 5| < \epsilon.\]
Draw two horizontal lines \(L + \epsilon , L - \epsilon\) and find the \(x\)-coordinate of intersection points with the graph of \(f(x)\).
In other words, set \(f(x) = L + \epsilon\) and \(f(x) - \epsilon\) to solve for \(x\), then the two solutions are \(x_1, x_2\). Then Choose \(\delta = \min(|x_1-a|, |x_2-a|)\).
Find a number \(\delta >0\) so that \(|\sqrt{x} -2| < 0.4\) whenever \(|x -4| < \delta\).
Set \(\sqrt{x} = 2+0.4\) and \(\sqrt{x} = 2 - 0.4\), then \(x_1 = 5.76, x_2 = 2.56\). Then choose \(\delta = \min(|5.76-4|, |2.56-4|) = 1.44.\)
Derivative at a number \[f'(a) = \lim_{h \to 0} \frac{f(a +h) - f(a)}{h}\]
Derivative as a function \[f'(x) = \lim_{h \to 0} \frac{f(x +h) - f(x)}{h}\]
The tangent line equation at the point \((a, f(a))\) is \[y - f(a) = f'(a)(x-a)\]
I suggest you compute \(f'(x)\) first, then plug in \(a\).
Derivative of \[f(x) = \frac{1}{x}\]
\[f'(x) = \lim_{h \to 0} \frac{\frac{1}{x+h} - \frac{1}{x}}{h} = \lim_{h \to 0} \frac{x - (x+h)}{x(x+h) h} = \lim_{h \to 0} \frac{-1}{x(x+h)} = \frac{-1}{x^2}\].
\[f(x) = \frac{3}{x+1}\]
\[f'(x) = \lim_{h \to 0} \frac{\frac{3}{x+h+1} - \frac{3}{x+1}}{h} = \lim_{h \to 0} \frac{3(x+1) - 3(x+h+1)}{(x+1)(x+h+1) h} = \lim_{h \to 0} \frac{-3}{(x+1)(x+h+1)} = \frac{-3}{(x+1)^2}\].
Find the tangent line of \(x^2\) at \((1,1)\).
\[f'(x) = \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h} = \lim_{h \to 0} \frac{2x h + h^2}{h} = \lim_{h \to 0} 2x + h = 2x\].
Plug in \(x = 1\), the slope is \(2\). So \[y -1 = 2(x-1)\] or \(y = 2x -1\) is the tangent line at \((1,1)\).