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Math's must do and practice question for 2nd internal Exam

Sreehari K
Limits & Continuity — Worked Solutions (step-by-step)

Limits & Continuity — Step-by-Step Solutions

I interpreted unclear lines as the most common textbook problems. If one of the expressions here is not the exact problem you expected, tell me which expression to change and I will produce the matching HTML immediately.

Exercise 1. Explain why the following limit does not exist:
\(\displaystyle \lim_{x\to 0}\frac{|x|}{x}\)
  1. Compute the right-hand limit: when \(x>0\), \(|x|=x\). So \[ \frac{|x|}{x}=\frac{x}{x}=1 \quad\text{for }x>0. \] Hence \(\displaystyle\lim_{x\to 0^+}\frac{|x|}{x}=1.\)
  2. Compute the left-hand limit: when \(x<0\), \(|x|=-x\). So \[ \frac{|x|}{x}=\frac{-x}{x}=-1 \quad\text{for }x<0. \] Hence \(\displaystyle\lim_{x\to 0^-}\frac{|x|}{x}=-1.\)
  3. Because the right-hand limit is \(1\) and the left-hand limit is \(-1\), the two one-sided limits are not equal. Therefore the two-sided limit \(\displaystyle\lim_{x\to 0}\frac{|x|}{x}\) does not exist.
Final answer: The limit does not exist because \(\lim_{x\to0^+}=1\) and \(\lim_{x\to0^-}=-1\).
Exercise 2. Explain why the following limit does not exist (most likely intended problem):
\(\displaystyle \lim_{x\to 1}\frac{|x-1|}{x-1}\)
  1. For \(x>1\): \(|x-1|=x-1\), so \(\dfrac{|x-1|}{x-1}=1\). Thus \(\displaystyle\lim_{x\to1^+}\dfrac{|x-1|}{x-1}=1.\)
  2. For \(x<1\): \(|x-1|=-(x-1)\), so \(\dfrac{|x-1|}{x-1}=-1\). Thus \(\displaystyle\lim_{x\to1^-}\dfrac{|x-1|}{x-1}=-1.\)
  3. Right and left limits differ, so the two-sided limit does not exist.
Final answer: The limit does not exist because the one-sided limits are \(1\) and \(-1\).
Exercise 3. (Interpreting your text) Suppose \(\displaystyle\lim_{x\to 1^-} f(x)=5\). Must \(f\) be defined at \(x=1\)? If it is, must \(f(1)=5\)? Can we conclude anything about the values of \(f\) at \(x=1\)?
  1. Definition reminder: A limit \(\lim_{x\to a} f(x)=L\) describes the behavior of \(f(x)\) as \(x\) approaches \(a\), not the value \(f(a)\) itself.
  2. So if \(\displaystyle\lim_{x\to 1^-} f(x)=5\):
    • \(f\) need not be defined at \(x=1\). The limit can exist even if \(f(1)\) is undefined.
    • If \(f\) is defined at 1, it is not necessary that \(f(1)=5\). The function could be defined with any value at \(x=1\) (for example, define \(f(1)=0\) or \(f(1)=100\)); that does not change the one-sided limit from the left.
    • The only conclusion we can make about \(f\) at \(x=1\) is that the left-hand values of \(f(x)\) approach \(5\). We cannot conclude the function's value at the point unless we are additionally told the function is continuous there (or given \(f(1)\)).
Final answer: \(f\) need not be defined at \(1\). If \(f(1)\) is defined it need not equal \(5\). We only know the left-hand limit equals \(5\).
Exercise 4. (Interpreted problem) Find the one-sided limit:
\(\displaystyle \lim_{x\to -2^+}\frac{x^3+2x}{x^2+5x+6}\)
  1. Factor the denominator: \(x^2+5x+6=(x+2)(x+3)\).
  2. Evaluate numerator at \(x=-2\): \( (-2)^3 + 2(-2) = -8 -4 = -12\). So the numerator is near \(-12\), a nonzero negative number, when \(x\) is near \(-2\).
  3. As \(x\to -2^+\) we have: \[ x+2 \to 0^+ \quad\text{(small positive)},\qquad x+3 \approx 1 \text{ (positive)}. \] Hence denominator \((x+2)(x+3)\to 0^+\) (small positive).
  4. So the fraction behaves like \(\dfrac{\text{(approx) } -12}{\text{small positive}}\) which goes to \(-\infty\).
Final answer: \(\displaystyle\lim_{x\to -2^+}\frac{x^3+2x}{x^2+5x+6}=-\infty.\)
Exercise 5. Suppose \(\displaystyle\lim_{x\to 0} f(x)=1\) and \(\displaystyle\lim_{x\to 0} g(x)=-5\). Evaluate
\(\displaystyle \lim_{x\to 0}\frac{2f(x)-g(x)}{\bigl(f(x)+7\bigr)^{2/3}}\)
  1. By the limit laws (sum/difference and constant multiple), \[ \lim_{x\to 0} \bigl(2f(x)-g(x)\bigr) = 2\lim_{x\to 0}f(x) - \lim_{x\to 0}g(x) = 2(1) - (-5) = 2 + 5 = 7. \]
  2. Compute the denominator limit: first \(\lim_{x\to0} (f(x)+7) = \lim f(x) + 7 = 1+7=8.\) The function \(u\mapsto u^{2/3}\) is continuous at \(u=8\), so \[ \lim_{x\to 0}\bigl(f(x)+7\bigr)^{2/3} = 8^{2/3}. \] Now compute \(8^{2/3} = (8^{1/3})^2 = 2^2 = 4.\)
  3. Apply the quotient rule for limits (denominator nonzero): \[ \lim_{x\to0}\frac{2f(x)-g(x)}{(f(x)+7)^{2/3}} = \frac{7}{4}. \]
Final answer: \(\displaystyle \lim_{x\to0}\frac{2f(x)-g(x)}{(f(x)+7)^{2/3}} = \frac{7}{4}.\)

Names of the rules used: (a) constant multiple rule, (b) sum/difference rule, (c) composition/continuity of the power function and the quotient rule for limits (denominator nonzero).

Exercise 6. Suppose \(5-2x^2\le f(x)\le 5-x^2\) for \(-1\le x\le 1\). Find \(\displaystyle\lim_{x\to 0} f(x)\).
  1. Evaluate the lower bound at \(x=0\): \(5-2(0)^2 = 5.\)
  2. Evaluate the upper bound at \(x=0\): \(5-(0)^2 = 5.\)
  3. So for \(x\) near \(0\), \[ 5 = 5-2x^2 \le f(x) \le 5-x^2 = 5. \] Both bounds approach \(5\) as \(x\to0\). By the Squeeze Theorem, \(\displaystyle\lim_{x\to0} f(x)=5.\)
Final answer: \(\displaystyle\lim_{x\to0} f(x)=5.\)
Exercise 7 & 8. Two typical limit implications (interpreting your lines):
  1. If \(\displaystyle\lim_{x\to2}\frac{f(x)}{x-2}=1\) (finite), then the denominator \(x-2\to 0\) while the quotient tends to a finite nonzero number. This forces the numerator to approach \(0\). Therefore \[ \lim_{x\to2} f(x)=0. \]
  2. If \(\displaystyle\lim_{x\to2}\frac{f(x)-5}{x-2}=3\) (finite), then again the denominator \(x-2\to 0\) and the quotient tends to a finite number, so the numerator must go to \(0\). Thus \[ \lim_{x\to2} (f(x)-5) = 0 \quad\Rightarrow\quad \lim_{x\to2} f(x)=5. \]
Final answers: From (a) \(\lim_{x\to2} f(x)=0\). From (b) \(\lim_{x\to2} f(x)=5\).

Selected Exercises — Continuity

1. For what points is the function \(\displaystyle y=\frac{1}{x-2}-3x\) continuous?
  1. \(\dfrac{1}{x-2}\) is continuous for all \(x\ne 2\) (denominator nonzero).
  2. \(-3x\) is continuous for all real \(x\).
  3. The sum of continuous functions is continuous on the intersection of domains. So the whole function is continuous for all real \(x\) except \(x=2\).
Final answer: Continuous for all \(x\in\mathbb{R}\) with \(x\ne 2\).
2. For what points is the function \(\displaystyle y=\frac{1}{|x|+1}\) continuous?
  1. \(|x|\) is continuous for all real \(x\).
  2. \(|x|+1\) is continuous and is never zero (always \(\ge 1\)).
  3. The reciprocal \(1/(|x|+1)\) is therefore continuous everywhere (since denominator never zero).
Final answer: Continuous for all real \(x\).
3. For what points is the function \(\displaystyle y=\frac{\sqrt{x^2+1}}{1+\sin^2 x}\) continuous?
  1. \(x\mapsto \sqrt{x^2+1}\) is continuous for all real \(x\) (inside \(\ge 1\)).
  2. \(x\mapsto 1+\sin^2 x\) is continuous and always \(\ge 1\), hence never zero.
    3. \item>The quotient of these is continuous for all real \(x\).
Final answer: Continuous for all real \(x\).
4. For which values of \(a\) and \(b\) is the piecewise function \[ g(x)= \begin{cases} ax+2b, & x\le 0,\\[4pt] x^2+3a-b, & 02 \end{cases} \] continuous for every \(x\)?
  1. Continuity at \(x=0\): left value at \(0\) is \(a\cdot 0 + 2b = 2b\). Right value (from middle piece at \(0^+\)) is \(0^2 + 3a - b = 3a-b\). Set equal: \[ 2b = 3a - b \quad\Rightarrow\quad 3b=3a \quad\Rightarrow\quad b=a. \]
  2. Continuity at \(x=2\): middle piece at \(2\) gives \(2^2 + 3a - b = 4 + 3a - b\). Right piece at \(2^+\) gives \(3(2)-5=6-5=1\). Set equal: \[ 4 + 3a - b = 1 \quad\Rightarrow\quad 3a - b = -3. \] Substitute \(b=a\): \(3a-a = -3 \Rightarrow 2a = -3 \Rightarrow a = -\tfrac{3}{2}.\)
  3. Then \(b=a = -\tfrac{3}{2}.\)
Final answer: \(a=b=-\dfrac{3}{2}.\)
5. For what value of \(a\) is \[ f(x)=\begin{cases} x^2-1, & x<3,\\[4pt] 2ax, & x\ge 3 \end{cases} \] continuous at every \(x\)?
  1. Only a potential discontinuity is at \(x=3\). Left limit at \(3\): \(3^2-1=9-1=8.\)
  2. Right value at \(3\): \(2a\cdot 3 = 6a.\)
  3. Set \(6a = 8\Rightarrow a=\dfrac{8}{6}=\dfrac{4}{3}.\)
Final answer: \(a=\dfrac{4}{3}.\)
6. (Ambiguous line in original text.) General remark about continuity & limits of composed trig functions:

If you have a limit \(\lim_{x\to c} h(x)\) where \(h(x)\) is composed from standard continuous functions (polynomials, roots, sines, cosines, tangents) and all inner values are defined at \(c\) (e.g. no division by zero, no \(\tan\) at its vertical asymptote), then the limit equals the direct substitution \(h(c)\). If you paste the exact expression you want evaluated here (the original line was garbled), I will compute it step by step and produce the exact numeric answer.

7. Roots of a cubic. Show that the equation \(x^3-15x+1=0\) has three solutions in \([-4,4]\).
  1. Define \(h(x)=x^3-15x+1\). Compute values at integer points inside \([-4,4]\):
  2. \(h(-4) = -64 + 60 + 1 = -3\) (negative).
  3. \(h(-3) = -27 + 45 + 1 = 19\) (positive). So there is a root in \((-4,-3)\) by the Intermediate Value Theorem (IVT).
  4. \(h(0)=1\) (positive) and \(h(1)=1-15+1=-13\) (negative). So there is a root in \((0,1)\) by IVT.
    5. \li>\(h(3)=27-45+1=-17\) (negative) and \(h(4)=64-60+1=5\) (positive). So there is a root in \((3,4)\) by IVT.
Final answer: By the IVT there are at least three distinct real roots, one in each interval \((-4,-3)\), \((0,1)\), and \((3,4)\). Hence three solutions lie in \([-4,4]\).
8. Removable discontinuity example. Give \(f(x)\) continuous everywhere except at \(x=2\) where it has a removable discontinuity. Explain how you know.
  1. Example: define \[ f(x)=\begin{cases}\dfrac{x^2-4}{x-2}, & x\ne 2,\\[6pt] \text{(not defined)}, & x=2.\end{cases} \]
  2. For \(x\ne 2\), \(\dfrac{x^2-4}{x-2}=\dfrac{(x-2)(x+2)}{x-2}=x+2\). So for \(x\ne 2\), \(f(x)=x+2\), which is continuous.
  3. \(\lim_{x\to 2} f(x)=\lim_{x\to 2}(x+2)=4\). But \(f(2)\) is not defined (or could be defined to some other number). Because the two-sided limit exists and equals \(4\) but \(f(2)\) is missing (or different), we have a removable discontinuity at \(x=2\).
  4. If we redefine \(f(2)=4\), the discontinuity is removed and the function becomes continuous everywhere.
Final answer: The function above has a removable discontinuity at \(x=2\); it is removable because the limit exists (equals 4) and we can redefine \(f(2)=4\) to make \(f\) continuous at 2.
9. Extend \(g(x)=\dfrac{x^2-9}{x-3}\) to be continuous at \(x=3\).
  1. Factor: \(\dfrac{x^2-9}{x-3}=\dfrac{(x-3)(x+3)}{x-3}=x+3\) for \(x\ne 3\).
  2. \(\lim_{x\to 3}g(x)=\lim_{x\to3}(x+3)=6.\)
  3. Define the extended function \(\displaystyle g(3)=6\). With that definition the function equals \(x+3\) for all \(x\) (and is continuous everywhere).
Final answer: Define \(g(3)=6\). Then \(g(x)\) is continuous at \(x=3\).

If any of the original expressions you posted were slightly different (for example a different algebraic expression in Exercises 1–8 or the trig limit in Continuity exercise 6), paste that exact expression and I will update the HTML instantly. Good luck on your exam — show the one-sided limit computations (as I did) to get full credit!

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