CBSE Class 11 Mathematics Limits and Derivatives Assignment Set 08

Exam Tip: Always evaluate both left-hand limit (LHL) and right-hand limit (RHL) separately when dealing with piecewise functions.

Question 2. If \(f(x) = \begin{cases} \frac{x - |x|}{x} ; x \neq 0 \\ 2 ; x = 0 \end{cases}\), show that \(\lim_{x \to 0} f(x)\) does not exist.
Answer: Here, for L.H.L. and R.H.L., \(f(x) = \frac{x - |x|}{x}\)
L.H.L. = \(\lim_{x \to 0^-} \left[\frac{x - |x|}{x}\right]\)
Put \(x = 0 - h = -h\) and \(h \to 0\)
\(\therefore\) L.H.L. = \(\lim_{h \to 0} \left(\frac{-h - |-h|}{-h}\right)\)
\(\Rightarrow\) L.H.L. = \(\lim_{h \to 0} \left(\frac{-h - h}{-h}\right) = \lim_{h \to 0} \left(\frac{-2h}{-h}\right)\)
\(\Rightarrow\) L.H.L. = \(\lim_{h \to 0} (2) = 2\)
Now, R.H.L. = \(\lim_{x \to 0^+} \left[\frac{x - |x|}{x}\right]\)
Put \(x = 0 + h = h\) and \(h \to 0\)
\(\therefore\) R.H.L. = \(\lim_{h \to 0} \left(\frac{h - |h|}{h}\right)\)
\(\Rightarrow\) R.H.L. = \(\lim_{h \to 0} \left(\frac{h - h}{h}\right) = \lim_{h \to 0} \left(\frac{0}{h}\right) = \lim_{h \to 0} (0) = 0\)
Since L.H.L. \(\neq\) R.H.L., because \(2 \neq 0\)
\(\therefore\) \(\lim_{x \to 0} f(x)\) does not exist.
In simple words: The limit does not exist because coming from the left gives 2, but coming from the right gives 0.

Exam Tip: Expand the modulus function carefully based on whether \(x\) is positive or negative.

Question 3. If \(f(x) = \begin{cases} 4x - 5 ; x \leq 2 \\ x - \pi ; x > 2 \end{cases}\), find the value of \(\pi\) if \(\lim_{x \to 2} f(x)\) exists.
Answer: For L.H.L., \(f(x) = 4x - 5\)
For R.H.L., \(f(x) = x - \pi\)
L.H.L. = \(\lim_{x \to 2^-} (4x - 5)\)
Put \(x = 2 - h\) and \(h \to 0\)
\(\Rightarrow\) L.H.L. = \(\lim_{h \to 0} (4(2 - h) - 5) = \lim_{h \to 0} (8 - 4h - 5) = \lim_{h \to 0} (3 - 4h) = 3\)
Now, R.H.L. = \(\lim_{x \to 2^+} (x - \pi)\)
Put \(x = 2 + h\) and \(h \to 0\)
\(\Rightarrow\) R.H.L. = \(\lim_{h \to 0} (2 + h - \pi) = 2 - \pi\)
Since \(\lim_{x \to 2} f(x)\) exists, L.H.L. = R.H.L.
\(\Rightarrow 3 = 2 - \pi\)
\(\Rightarrow \pi = 2 - 3 = -1\)
\(\therefore\) \(\pi = -1\)
In simple words: For a limit to exist at a point, the left-hand limit must equal the right-hand limit. Setting them equal helps find the unknown constant.

Exam Tip: Note that here the Greek letter \(\pi\) is simply used as a constant variable name in the function definition.

Question 4. Show that \(\lim_{x \to 0} \left(\frac{e^{1/x} - 1}{e^{1/x} + 1}\right)\) does not exist.
Answer: For L.H.L., \(f(x) = \frac{e^{1/x} - 1}{e^{1/x} + 1}\)
L.H.L. = \(\lim_{x \to 0^-} \left(\frac{e^{1/x} - 1}{e^{1/x} + 1}\right)\)
Put \(x = 0 - h = -h\) and \(h \to 0\)
\(\therefore\) L.H.L. = \(\lim_{h \to 0} \left(\frac{e^{-1/h} - 1}{e^{-1/h} + 1}\right)\)
Substitute \(h = 0\), noting that \(e^{-\infty} = 0\)
\(\Rightarrow\) L.H.L. = \(\frac{0 - 1}{0 + 1} = -1\)
Now, R.H.L. = \(\lim_{x \to 0^+} \left(\frac{e^{1/x} - 1}{e^{1/x} + 1}\right)\)
Put \(x = 0 + h = h\) and \(h \to 0\)
\(\therefore\) R.H.L. = \(\lim_{h \to 0} \left(\frac{e^{1/h} - 1}{e^{1/h} + 1}\right)\)
Dividing numerator and denominator by \(e^{1/h}\):
\(\Rightarrow\) R.H.L. = \(\lim_{h \to 0} \left(\frac{1 - e^{-1/h}}{1 + e^{-1/h}}\right)\)
Substitute \(h = 0\):
\(\Rightarrow\) R.H.L. = \(\frac{1 - 0}{1 + 0} = 1\)
Since L.H.L. \(\neq\) R.H.L. (\(-1 \neq 1\)), the limit does not exist.
In simple words: The function approaches -1 from the left and 1 from the right, so the overall limit does not exist.

Exam Tip: For exponential terms with reciprocal variables, evaluate limits at zero by using the standard property \(e^{-\infty} = 0\).

Question 5. If \(f(x) = \begin{cases} a + bx ; x < 1 \\ 4 ; x = 1 \\ b - ax ; x > 1 \end{cases}\) and if \(\lim_{x \to 1} f(x) = f(1)\), what are possible values of \(a\) and \(b\)?
Answer: Given \(f(1) = 4\) and \(\lim_{x \to 1} f(x) = f(1)\)
\(\Rightarrow \lim_{x \to 1} f(x) = 4\)
This implies that L.H.L. = R.H.L. = 4
L.H.L. = \(\lim_{x \to 1^-} (a + bx) = a + b\)
R.H.L. = \(\lim_{x \to 1^+} (b - ax) = b - a\)
Therefore, \(a + b = 4\) and \(b - a = 4\)
Adding both equations:
\((a + b) + (b - a) = 4 + 4\)
\(2b = 8\)
\(b = 4\)
Substituting \(b = 4\) into \(a + b = 4\):
\(a + 4 = 4\)
\(a = 0\)
Thus, \(a = 0\) and \(b = 4\).
In simple words: By equating the left-hand and right-hand limits to the given function value 4, we form a system of linear equations to solve for \(a\) and \(b\).

Exam Tip: Continuity at a point requires the limit of the function to equal the value of the function at that point.

Question 6. If \(f(x) = \begin{cases} mx^2 + n ; x < 0 \\ nx + m ; 0 \leq x \leq 1 \\ nx^3 + m ; x > 1 \end{cases}\), for what integers \(m\) and \(n\) do the limits \(\lim_{x \to 0} f(x)\) and \(\lim_{x \to 1} f(x)\) exist?
Answer: For \(\lim_{x \to 0} f(x)\) to exist, L.H.L. = R.H.L. at \(x = 0\):
\(\lim_{x \to 0^-} (mx^2 + n) = \lim_{x \to 0^+} (nx + m)\)
\(\Rightarrow m(0)^2 + n = n(0) + m\)
\(\Rightarrow n = m \quad \text{--- (i)}\)
For \(\lim_{x \to 1} f(x)\) to exist, L.H.L. = R.H.L. at \(x = 1\):
\(\lim_{x \to 1^-} (nx + m) = \lim_{x \to 1^+} (nx^3 + m)\)
\(\Rightarrow n(1) + m = n(1)^3 + m\)
\(\Rightarrow n + m = n + m\)
This is an identity which is always true for any real numbers \(m\) and \(n\).
From equation (i), we have \(m = n\).
Thus, \(m\) and \(n\) can be any integers such that \(m = n\).
In simple words: Equating limits at the boundary points gives conditions relating \(m\) and \(n\). Here, both limits exist as long as \(m\) and \(n\) are equal integers.

Exam Tip: Check each transition point independently by setting the left-side expression equal to the right-side expression.

Question 7. If \(f(x) = \begin{cases} |x| + 1 ; x < 0 \\ 0 ; x = 0 \\ |x| - 1 ; x > 0 \end{cases}\), for what value(s) of \(a\) does \(\lim_{x \to a} f(x)\) exist?
Answer: Let us check the limit at \(x = 0\):
L.H.L. = \(\lim_{x \to 0^-} (|x| + 1) = \lim_{h \to 0} (|-h| + 1) = 0 + 1 = 1\)
R.H.L. = \(\lim_{x \to 0^+} (|x| - 1) = \lim_{h \to 0} (|-h| - 1) = 0 - 1 = -1\)
Since L.H.L. \(\neq\) R.H.L. at \(x = 0\) (\(1 \neq -1\)), \(\lim_{x \to 0} f(x)\) does not exist.
For any other real number \(a \neq 0\) in the domain, the function is continuous and the limit exists.
Therefore, \(a\) can be any real number except \(0\), i.e., \(a \in R - \{0\}\).
In simple words: The limit exists everywhere except at \(x = 0\), where the function makes a sudden jump from 1 to -1.

Exam Tip: Test points where the function definition changes or where absolute values can create discontinuities.

Question 8. If \(f(x) = \begin{cases} 2x + 3 ; x \leq 0 \\ 3(x + 1) ; x > 0 \end{cases}\), evaluate \(\lim_{x \to 1} f(x)\).
Answer: We need to evaluate \(\lim_{x \to 1} f(x)\). Since \(x = 1 > 0\), we use the function definition for \(x > 0\), which is \(f(x) = 3(x + 1)\).
L.H.L. = \(\lim_{x \to 1^-} 3(x + 1)\)
Put \(x = 1 - h\) and \(h \to 0\)
\(\therefore\) L.H.L. = \(\lim_{h \to 0} 3(1 - h + 1) = 3(2) = 6\)
R.H.L. = \(\lim_{x \to 1^+} 3(x + 1)\)
Put \(x = 1 + h\) and \(h \to 0\)
\(\therefore\) R.H.L. = \(\lim_{h \to 0} 3(1 + h + 1) = 3(2) = 6\)
Since L.H.L. = R.H.L. = 6
\(\therefore\) \(\lim_{x \to 1} f(x)\) exists and \(\lim_{x \to 1} f(x) = 6\)
In simple words: Since 1 is greater than 0, we only need to use the second branch of the function, which is continuous at \(x = 1\).

Exam Tip: Always check which interval the target point belongs to before choosing the function branch.

Question 9. Let \(a_1, a_2, a_3, \dots, a_n\) be any real numbers and \(f(x) = (x - a_1)(x - a_2)(x - a_3)\dots(x - a_n)\). What is \(\lim_{x \to a_1} f(x)\)? Also compute \(\lim_{x \to a} f(x)\).
Answer: For \(\lim_{x \to a_1} f(x)\):
\(\lim_{x \to a_1} [(x - a_1)(x - a_2)(x - a_3)\dots(x - a_n)]\)
\(= (a_1 - a_1)(a_1 - a_2)(a_1 - a_3)\dots(a_1 - a_n)\)
\(= 0 \cdot (a_1 - a_2)(a_1 - a_3)\dots(a_1 - a_n) = 0\)
For \(\lim_{x \to a} f(x)\):
\(\lim_{x \to a} [(x - a_1)(x - a_2)\dots(x - a_n)]\)
\(= (a - a_1)(a - a_2)\dots(a - a_n)\)
In simple words: Direct substitution evaluates the product to zero when \(x = a_1\) because one of the factors becomes zero.

Exam Tip: Polynomial functions are continuous everywhere, so limits can be evaluated by direct substitution.

Type: 2 Factorize
Formula: \(a^2 - b^2, a^3 - b^3, a^4 - b^4\), quadratic equation, cubic (hit and trial) L.C.M.

Question 10. Evaluate: \(\lim_{x \to 1} \left[ \frac{x - 2}{x^2 - x} - \frac{1}{x^3 - 3x^2 + 2x} \right]\)
Answer: We have \(\lim_{x \to 1} \left[ \frac{x - 2}{x^2 - x} - \frac{1}{x^3 - 3x^2 + 2x} \right]\)
Factorizing denominators:
\(x^2 - x = x(x - 1)\)
\(x^3 - 3x^2 + 2x = x(x^2 - 3x + 2) = x(x - 1)(x - 2)\)
Substituting the factors back into the expression:
\(= \lim_{x \to 1} \left[ \frac{x - 2}{x(x - 1)} - \frac{1}{x(x - 1)(x - 2)} \right]\)
Taking the L.C.M. as \(x(x - 1)(x - 2)\):
\(= \lim_{x \to 1} \left[ \frac{(x - 2)^2 - 1}{x(x - 1)(x - 2)} \right]\)
Expanding the numerator:
\(= \lim_{x \to 1} \left[ \frac{x^2 - 4x + 4 - 1}{x(x - 1)(x - 2)} \right]\)
\(= \lim_{x \to 1} \left[ \frac{x^2 - 4x + 3}{x(x - 1)(x - 2)} \right]\)
Factorizing the quadratic numerator \(x^2 - 4x + 3 = (x - 3)(x - 1)\):
\(= \lim_{x \to 1} \left[ \frac{(x - 3)(x - 1)}{x(x - 1)(x - 2)} \right]\)
Canceling the common term \((x - 1)\):
\(= \lim_{x \to 1} \left[ \frac{x - 3}{x(x - 2)} \right]\)
Substituting \(x = 1\):
\(= \frac{1 - 3}{1(1 - 2)} = \frac{-2}{-1} = 2\)
In simple words: Combine the rational expressions by finding a common denominator, factor all polynomials, cancel common factors, and substitute \(x = 1\).

Exam Tip: Always factor denominators completely before attempting to find a common denominator in rational limit problems.