CBSE Class 11 Mathematics Limits And Derivatives Worksheet Set 05

CBSE Class 11 Mathematics Worksheet - Limits and Derivatives. The questions in the worksheets have been specifically designed by best teachers so that the students can practise them to clear their concepts and get better marks in tests and examinations. Students can download these worksheets and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

Question. If \( f(x) = \begin{cases} 5x - 4 & ; \ 0 < x < 1 \\ 4x^3 - 3x & ; \ 1 < x < 2 \end{cases} \)
Evaluate \( \lim_{x\to1} f(x) \)
Answer: For L.H.L. : \( f(x) = 5x - 4 \)
For R.H.L : \( f(x) = 4x^3 - 3x \)
L.H.L. \( = \lim_{x\to1^-} (5x - 4) \)
Put \( x = 1 - h \) & \( h \to 0 \)
\( \therefore \) L.H.L. \( = \lim_{h\to0} (5(1 - h) - 4) \)
\( \implies \) L.H.L \( = 5 - 4 = 1 \)
\( \therefore \) L.H.L. \( = 1 \)
Now, R.H.L. \( = \lim_{x\to1^+} (4x^3 - 3x) \)
Put \( x = 1 + h \) & \( h \to 0 \)
\( \therefore \) R.H.L. \( = \lim_{h\to0} [4(1 + h)^3 - 3(1 + h)] \)
\( \implies \) R.H.L \( = 4(1)^3 - 3(1) = 1 \)
\( \therefore \) R.H.L. \( = 1 \)
Since, L.H.L. = R.H.L. = 1
\( \therefore \lim_{x\to1} f(x) \) Exists & \( \lim_{x\to1} f(x) = 1 \) ans.

Question. If \( f(x) = \begin{cases} \frac{x-|x|}{x} & ; \ x \neq 0 \\ 2 & ; \ x = 0 \end{cases} \)
Show that \( \lim_{x\to0} f(x) \) does not exists.
Answer: Here, for L.H.L. & R.H.L.
\( f(x) = \frac{x - |x|}{x} \)
L.H.L. \( = \lim_{x\to0^-} \left[ \frac{x-|x|}{x} \right] \)
Put \( x = 0 - h = -h \) & \( h \to 0 \)
\( \therefore \) L.H.L. \( = \lim_{h\to0} \left( \frac{-h-|-h|}{-h} \right) \)
\( \implies \) L.H.L. \( = \lim_{h\to0} \left( \frac{-h-h}{-h} \right) = \lim_{h\to0} \left( \frac{-2h}{-h} \right) \)
\( \implies \lim_{h\to0} (2) \)
\( \therefore \) L.H.L. \( = 2 \)
Now, R.H.L. \( = \lim_{x\to0^+} \left[ \frac{x-|x|}{x} \right] \)
Put \( x = 0 + h = h \) & \( h \to 0 \)
\( \therefore \) R.H.L. \( = \lim_{h\to0} \left( \frac{h-|h|}{h} \right) = \lim_{h\to0} \left( \frac{h-h}{h} \right) \)
\( \implies \) R.H.L. \( = \lim_{h\to0} \left( \frac{0}{h} \right) = \lim_{h\to0} (0) = 0 \)
\( \therefore \) R.H.L. \( = 0 \)
Clearly L.H.L. \( \neq \) R.H.L.
\( \therefore \lim_{x\to0} f(x) \) does not exists ans.

Question. \( f(x) = \begin{cases} 4x - 5 & ; \ x \le 2 \\ x - a & ; \ x > 2 \end{cases} \)
Find value of \( a \) if \( \lim_{x\to2} f(x) \) exists.
Answer: For L.H.L. \( f(x) = 4x - 5 \)
For R.H.L. \( f(x) = x - a \)
L.H.L. \( = \lim_{x\to2^-} (4x - 5) \)
Put \( x = 2 - h \) & \( h \to 0 \)
\( \implies \) L.H.L. \( = \lim_{h\to0} (4(2 - h) - 5) = \lim_{h\to0} (8 - 5) \)
\( \implies \) L.H.L. \( = \lim_{h\to0} (3) \)
\( \implies \) L.H.L. \( = 3 \)
Now, R.H.L. \( = \lim_{x\to2^+} (x - a) \)
Put \( x = 2 + h \) and \( h \to 0 \)
\( \implies \) R.H.L. \( = \lim_{h\to0} (2 + h - a) = \lim_{h\to0} (2 - a) \)
\( \therefore \) R.H.L. \( = 2 - a \)
Since, L.H.L. = R.H.L.
\( \implies 3 = 2 - a \implies a = -1 \) ans.

Question. Show that \( \lim_{x\to0} \left( \frac{e^{1/x}-1}{e^{1/x}+1} \right) \) does not exists.
Answer: For L.H.L. & R.H.: \( f(x) = \left( \frac{e^{1/x}-1}{e^{1/x}+1} \right) \)
L.H.L. \( = \lim_{x\to0^-} \left( \frac{e^{1/x}-1}{e^{1/x}+1} \right) \)
Put \( x = 0 - h = -h \) & \( h \to 0 \)
\( \therefore \) L.H.L. \( = \lim_{h\to0} \left( \frac{e^{-1/h}-1}{e^{-1/h}+1} \right) \)
Put directly \( h = 0 \)
\( \implies \) L.H.L \( = \frac{e^{-\infty}-1}{e^{-\infty}+1} = \frac{0-1}{0+1} \) \quad \( e^{-\infty} = 0 \)
\( \implies \) L.H.L. \( = -1 \)
Now, R.H.L. \( = \lim_{x\to0^+} \left( \frac{e^{1/x}-1}{e^{1/x}+1} \right) \)
Put \( x = 0 + h = h \) & \( h \to 0 \)
\( \therefore \) R.H.L. \( = \lim_{h\to0} \left( \frac{e^{1/h}-1}{e^{1/h}+1} \right) \)
(don't put directly \( h = 0 \)) \( \frac{\infty}{\infty} \) form
\( \implies \) R.H.L. \( = \lim_{h\to0} \left( \frac{1-\frac{1}{e^{1/h}}}{1+\frac{1}{e^{1/h}}} \right) \) divide by \( e^{1/h} \)
\( \implies \) R.H.L. \( = \lim_{h\to0} \left( \frac{1-e^{-1/h}}{1+e^{-1/h}} \right) \)
Put \( h = 0 \)
\( \implies \) R.H.L. \( = \lim_{h\to0} \left( \frac{1-e^{-\infty}}{1+e^{-\infty}} \right) = \frac{1-0}{1+0} \) \quad \( e^{-\infty} = 0 \)
\( \implies \) R.H.L. \( = 1 \)
Since, L.H.L. \( \neq \) R.H.L.

\( \dots \lim_{x\to0} f(x) \) does not Exists ans.

Question. \( f(x) = \begin{cases} a + bx & ; \ x < 1 \\ 4 & ; \ x = 1 \\ b - ax & ; \ x > 1 \end{cases} \)
and if \( \lim_{x\to1} f(x) = f(1) \) , what are possible values of \( a \) & \( b \)?
Answer: For L.H.L. \( f(x) = a + bx \)
For R.H.L. \( f(x) = b - ax \)
and \( f(1) = 4 \) (when \( x = 1 \); \( f(x) = 4 \))
given, \( \lim_{x\to1} f(x) = f(1) \)
\( \implies \lim_{x\to1} f(x) = 4 \)
\( \implies \) L.H.L. = R.H.L. = 4
\( \implies \lim_{x\to1^-} (a + bx) = \lim_{x\to1^+} (b - ax) = 4 \)
Put \( x = 1 - h \) \quad Put \( x = 1 + h \)
\( \& \ h \to 0 \) \quad \( \& \ h \to 0 \)
\( \implies \lim_{h\to0} (a + b(1 - h)) = \lim_{h\to0} (b - a(1 + h)) = 4 \)
\( \implies a + b = b - a = 4 \)
\( \implies a + b = 4 \ \& \ b - a = 4 \)
Solving we get \( a = 0 \ \& \ b = 4 \) ans.

Question. \( f(x) = \begin{cases} mx^2 + n & ; \ x < 0 \\ nx + m & ; \ 0 \le x \le 1 \\ nx^3 + m & ; \ x > 1 \end{cases} \)
For what integers \( m \) and \( n \) does the \( \lim_{x\to0} f(x) \) and \( \lim_{x\to1} f(x) \) exists.
Answer: Given that \( \lim_{x\to0} f(x) \) exists
For L.H.L. \( f(x) = mx^2 + n \)
L.H.L = R.H.L.
\( \implies \lim_{x\to0^-} (mx^2 + n) = \lim_{x\to0^+} (nx + m) \)
Put \( x = 0 - h = -h \) \quad Put \( x = 0 + h = h \)
\( \& \ h \to 0 \) \quad \( \& \ h \to 0 \)
\( \therefore \lim_{h\to0} (m(-h)^2 + n) = \lim_{h\to0} (n(h) + m) \)
\( \implies 0 + n = 0 + m \)
\( \implies m = n \) ............ (i)
Given that \( \lim_{x\to1} f(x) \) exists
For L.H.L. \( f(x) = nx + m \)
For R.H.L. \( f(x) = nx^3 + m \)
L.H.L. = R.H.L.
\( \implies \lim_{x\to1^-} (nx + m) = \lim_{x\to1^+} (nx^3 + m) \)
Put \( x = 1 - h \) \quad Put \( x = 1 + h \)
\( \& \ h \to 0 \) \quad \( \& \ h \to 0 \)
\( \implies \lim_{h\to0} (n(1 - h) + m) = \lim_{h\to0} (n(1 + h)^3 + m) \)
(put directly \( h = 0 \))
\( \implies n + m = n + m \) ............ (ii)
From (i) & (ii)
\( m \) and \( n \) can be any integers such that \( m = n \) ans.

Question. \( f(x) = \begin{cases} |x| + 1 & ; \ x < 0 \\ 0 & ; \ x = 0 \\ |x| - 1 & ; \ x > 0 \end{cases} \)
For what value(s) of \( a \) does the \( \lim_{x\to a} f(x) \) exists.
Answer: For L.H.L. \( f(x) = |x| + 1 \)
For R.H.L. \( f(x) = |x| - 1 \)
L.H.L. \( = \lim_{x\to0^-} (|x| + 1) \)
Put \( x = 0 - h = -h \) & \( h \to 0 \)
\( \therefore \) L.H.L. \( = \lim_{h\to0} (|-h| + 1) = \lim_{h\to0} (h + 1) = 0 + 1 \)
\( \implies \) L.H.L. \( = 1 \)
Now, R.H.L. \( = \lim_{x\to0^+} (|x| - 1) \)
Put \( x = 0 + h = h \) & \( h \to 0 \)
\( \therefore \) R.H.L. \( = \lim_{h\to0} (|h| - 1) = \lim_{h\to0} (h - 1) = 0 - 1 \)
\( \implies \) R.H.L. \( = -1 \)
Since L.H.L \( \neq \) R.H.L.
\( \therefore \lim_{x\to0} f(x) \) Does not exist. ............. (i)
But we are given, \( \lim_{x\to a} f(x) \) exists. ............. (ii)
From (i) & (ii)
We conclude that \( a \) can be any real no. except \( a = 0 \)
\( \therefore a \in \mathbb{R} - \{0\} \) ans.

Question. \( f(x) = \begin{cases} 2x + 3 & ; \ x \le 0 \\ 3(x + 1) & ; \ x > 0 \end{cases} \)
Evaluate \( \lim_{x\to1} f(x) \)
Answer: For L.H.L. \( f(x) = 3(x + 1) \)
Also for R.H.L. \( f(x) = 3(x + 1) \)
L.H.L. \( = \lim_{x\to1^-} 3(x + 1) \)
Put \( x = 1 - h \) & \( h \to 0 \)
\( \therefore \) L.H.L. \( = \lim_{h\to0} (3(1 - h + 1)) = 3(2) = 6 \)
\( \implies \) L.H.L. \( = 6 \)
Now, R.H.L. \( = \lim_{x\to1^+} (3(x + 1)) \)
Put \( x = 1 + h \) & \( h \to 0 \)
\( \therefore \) R.H.L. \( = \lim_{h\to0} (3(1 + h + 1)) = 3(2) = 6 \)
\( \implies \) R.H.L. \( = 6 \)
Since L.H.L. = R.H.L.
\( \therefore \lim_{x\to1} f(x) \) Exist and \( \lim_{x\to1} f(x) = 6 \) ans.

Question. \( a_1, a_2, a_3 \dots a_n \) are any real numbers \( f(x) = (x - a_1)(x - a_2)(x - a_3)\dots(x - a_n) \).
What is \( \lim_{x\to a_1} f(x) \)? Also compute \( \lim_{x\to a} f(x) \).
Answer: \( \lim_{x\to a_1} f(x) = \lim_{x\to a_1} [(x - a_1)(x - a_2)(x - a_3)\dots(x - a_n)] \)
\( = (a_1 - a_1)(a_1 - a_2)(a_1 - a_3)\dots(a_1 - a_n) \)
\( = 0(a_1 - a_2)(a_1 - a_3)\dots(a_1 - a_n) \)
\( = 0 \) ans.
\( \lim_{x\to a} f(x) = \lim_{x\to a} [(x - a_1)(x - a_2)\dots(x - a_n)] \)
\( = (a - a_1)(a - a_2)\dots(a - a_n) \) ans.

Question. Evaluate : \( \lim_{x\to1} \left[ \frac{x-2}{x^2-x} - \frac{1}{x^3-3x^2+2x} \right] \)
Answer: We have \( \lim_{x\to1} \left[ \frac{x-2}{x^2-x} - \frac{1}{x^3-3x^2+2x} \right] \)
\( = \lim_{x\to1} \left[ \frac{x-2}{x(x-1)} - \frac{1}{x(x^2-3x+2)} \right] \)
\( = \lim_{x\to1} \left[ \frac{x-2}{x(x-1)} - \frac{1}{x(x-1)(x-2)} \right] \)
\( = \lim_{x\to1} \left[ \frac{(x-2)^2-1}{x(x-1)(x-2)} \right] \)
\( = \lim_{x\to1} \left[ \frac{x^2-4x+4-1}{x(x-1)(x-2)} \right] \)
\( = \lim_{x\to1} \left[ \frac{(x-3)(x-1)}{x(x-1)(x-2)} \right] \)
\( = \lim_{x\to1} \left[ \frac{x-3}{x(x-2)} \right] \)
\( = \frac{1-3}{1(1-2)} = \frac{-2}{-1} = 2 \) ans.