Limit and Continuity Exercise: 16.4 Class 11 Basic Mathematics Solution [NEB UPDATED]

Exercise 16.4 

1.

(i)

Solution:

f(x) = x²

Left hand limit at x = 4 is

= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 4 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 4 – x² = (4)² = 16.

Right hand limit at x = 4 is

= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 4 + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 4 + x² = (4)² = 16.

f(4) = (4)² = 16.

So, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 4 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 4 + f(x) = f(4)

So, f(x) is continuous at x = 4.

 

(ii)
Solution:

f(x) = 2 – 3x²

Left hand limit at x = 0 is

= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\ \to\end{array}$ 0 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 – (2 – 3x²) = 2 – 0 = 2.

Right hand limit at x = 0 is

= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 + (2 – 3x²) = 2 – 0 = 2.

f(0) = 2 – 3 * 0 = 2.

So, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 + f(x) = f(0)

So, f(x) is continuous at x = 0.

 

(iii)

Solution:

f(x) = 3x² – 2x + 4.

Left hand limit at x = 1 is

= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 – (3x² – 2x + 4) = 3 * 1 – 2 * 1 + 4 = 5.

Right hand limit at x = 1 is

= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 + (3x² – 2x + 4) = 3 * 1 – 2 * 1 + 4 = 5.

f(1) = 3 * (1)² – 2 * 1 + 4 = 5.

So, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 + f(x) = f(1)

So, f(x) is continuous at x = 1.

 

(iv)

Solution:

f(x) = $\frac{1}{{2{\rm{x}}}}$.

f(0) = $\frac{1}{{2.0}}$ = $\frac{1}{0}$ = does not exist.

Hence, f(x) is discontinuous at x = 0.

 

(v)

Solution:

f(0) = $\frac{1}{{{\rm{x}} - 2}}$

Let x = a ≠ 2.

= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a –${\rm{\: }}\frac{1}{{{\rm{x}} - 2}}$ = $\frac{1}{{{\rm{a}} - 2}}$

= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a + $\frac{1}{{{\rm{x}} - 2}}{\rm{\: }}$= $\frac{1}{{{\rm{a}} - 2}}$.

f(a) = $\frac{1}{{{\rm{a}} - 2}}$.

So, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a + f(x) = f(a)

So, f(x) is continuous at x = a ≠ 2.

Note: since a ≠ 2, so a – 2 ≠ 0 and hence $\frac{1}{{{\rm{a}} - 2}}$ is a finite number.

 

(vi)

f(x) = $\frac{1}{{3{\rm{x}}}}$

Let x = a ≠ 0.

= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a${\rm{\: }}\frac{1}{{3{\rm{x}}}}$ = $\frac{1}{{3{\rm{a}}}}$

= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a + $\frac{1}{{3{\rm{x}}}}{\rm{\: }}$= $\frac{1}{{3{\rm{a}}}}$.

f(a) = $\frac{1}{{3{\rm{a}}}}$.

So, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ a + f(x) = f(a)

So,f(x) is continuous at x = a ≠ 0.

 

(vii)
Solution:

f(x) = $\frac{1}{{1 - {\rm{x}}}}$

f(1) = $\frac{1}{{1 - 1}}$ = $\frac{1}{0}$ which does  not exist.

So, f(x) is discontinuous at x = 3.

 

 

(ix)

Solution:

f(x) = $\frac{{{{\rm{x}}^2} - 9}}{{{\rm{x}} - 3}}$

RHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 $\frac{{{{\rm{x}}^2} - 9}}{{{\rm{x}} - 3}}$

= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 $\frac{{\left( {{\rm{x}} + 3} \right)\left( {{\rm{x}} - 3} \right)}}{{{\rm{x}} - 3}}$ = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 (x + 3) = 3 + 3 = 6.

LHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 $\frac{{{{\rm{x}}^2} - 9}}{{{\rm{x}} - 3}}$.

= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 – $\frac{{\left( {{\rm{x}} + 3} \right)\left( {{\rm{x}} - 3} \right)}}{{{\rm{x}} - 3}}$ = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 (x + 3) = 3 + 3 = 6.

f(3) = $\frac{{9 - 9}}{{3 - 3}}$ = $\frac{0}{0}$.

So, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 – f(x) ≠ f(3)

So, f(x) is continuous at x = 3.

Note: since a ≠ 2, so a – 2 ≠ 0 and hence $\frac{1}{{{\rm{a}} - 2}}$ is a finite number.

 

(x)

Solution:

f(x) = $\frac{{{{\rm{x}}^2} - 16}}{{{\rm{x}} - 4}}$

LHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 4 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 4 $\frac{{{{\rm{x}}^2} - 16}}{{{\rm{x}} - 4}}$

= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 4 – $\frac{{\left( {{\rm{x}} + 4} \right)\left( {{\rm{x}} - 4} \right)}}{{{\rm{x}} - 4}}$ = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 4 – (x + 4) = 4 + 4 = 8.

LHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 4 + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 4 + $\frac{{\left( {{\rm{x}} + 4} \right)\left( {{\rm{x}} - 4} \right)}}{{{\rm{x}} - 4}}$ = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 4 (x + 4) +  4 = 8.

f(x) = $\frac{{{{\left( 4 \right)}^2} - 16}}{{4 - 4}}$ = $\frac{0}{0}$.

So, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 4 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 4 + f(x) ≠ f(0)

So, f(x) is discontinuous at x = 4.

Note: since a ≠ 2, so a – 2 ≠ 0 and hence $\frac{1}{{{\rm{a}} - 2}}$ is a finite number.

 

2.

(i)

Solution:

For  x ≤ 2, f(x) = 2 – x².

LHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 – (2 – x²) = 2 – 4 = –2.

For x > 2, f(x) = x – 4

RHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 + (x – 4) = 2 – 4 = –2.

For, x = 2, f(x) = 2 – x²

f(2) = 2 – (2)² = –2.

x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 + f(x) = f(2)

So, f(x) is continuous at x = 2.

 

(ii)

Solution:

For x < 2, f(x) = 2x² + 1.

LHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 – (2x² + 1) = 2 *4 + 1 = 9.

For x > 2, f(x) = 4x + 1.

RHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 + (4x + 1) = 4 * 1 + 1 = 9.

For, x = 2, f(x) = 2x² + 1.

f(2) = 2 * (2)² + 1 = 9.

x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 + f(x) = f(2)

So, f(x) is continuous at x = 2.

 

(iii)

Solution:

For x < 3, f(x) = 2x.

LHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 – 2x = 2 *3 = 6.

For x > 3, f(x) = 3x – 3.

RHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 + (3x – 3) = 3 * 3 – 3 = 6.

For, x = 3, f(x) = 2x

f(3) = 2 * 3 = 6.

x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 + f(x) = f(3)

So, f(x) is continuous at x = 3.

 

(iv)

Solution:

For x < 1, f(x) = 2x + 1.

LHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 – (2x + 1) = 2 *1 + 1 = 3.

For x > 1, f(x) = 3x.

RHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 + 3x = 3 * 1 = 3.

For, x = 1, f(1) = 2.

x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 + f(x) ≠ f(1)

So, f(x) is discontinuous at x = 1.

 

3

(i)

Solution:

LHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 5^– f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 5^– (x² + 2) = 25 + 2  = 27.

RHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 5⁺ f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 5⁺(3x + 12) = 3.5 + 12 = 27.

Functional value = f(5) = 20.

Here, LHL = RHL ≠ f(5).

Therefore, given function is removable discontinuous at x = 5.

But, the given function can be made continuous by redefining as:

 

f(x) = $\{ \begin{array}{*{20}{c}}{{{\rm{x}}^2} + 2}&{{\rm{for\: x}} < 5}\\{27}&{{\rm{for\: x}} = 5}\\{3{\rm{x}} + 12}&{{\rm{for\: x}} > 5}\end{array}$

 

3

(ii)

Solution:

For x < 2, f(x) = 2x – 3.

LHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\ \to\end{array}$ 2 – (2x – 3) = 2 *2 – 3 = 1.

For x > 2, f(x) = 3x – 5.

RHL = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 + (3x – 5) = 3 * 2 – 5 = 1.

f (2) = 2.

x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 + f(x) ≠ f(2)

So, f(x) is discontinuous at x = 2.

So, the given function will be continuous if f(x) is redefined as follows:

f(x) = $\{ \begin{array}{*{20}{c}}{2{\rm{x}} - 3}&{{\rm{for\: x}} < 2}\\1&{{\rm{for\: x}} = 2}\\{3{\rm{x}} - 5}&{{\rm{for\: x}} > 2}\end{array}{\rm{\: \: }}$

4.

(i)

Solution:

For x < 2, f(x) = kx + 3.

= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 (3x – 1) = 3.2 – 1 = 5.

Being continuous,

= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 – f(x)

2k + 3 = 5.

So, k = 1.

 

 

 

(ii)

Solution:

For x > 3 and x < 3, f(x) = $\frac{{2{{\rm{x}}^2} - 19}}{{{\rm{x}} - 13}}$.

= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 $\frac{{2{{\rm{x}}^2} - 18}}{{{\rm{x}} - 3}}$ = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 3 $\frac{{2\left( {{\rm{x}} - 3} \right)\left( {{\rm{x}} + 3} \right)}}{{{\rm{x}} - 3}}$.

= 2(3 + 3) = 12.

For, x = 3, f(x) = k, i.e. f(3) = k.

Being continuous,

12 = k

So, k = 12.