Limit and Contiunity Exercise: 16.3 Class 11 Basic Mathematics Solution [NEB UPDATED]

Exercise: 16.3

1.

a.

Solution:
For x ≥ 0, f(x) = x + 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 + (x + 2) = 0 + 2 = 0,

For x < 0, f(x) = 4x + 2

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 – (4x + 2) = 0 + 2 = 2.

= 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)

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

 

b.

Solution:
For x ≥ 1, f(x) = 3x + 2

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 + (3x + 2) = 3*1 + 2 = 5,

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

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 – (2x) = 2 * 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)

So, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 1 f(x) does not exist.

 

c.

Solution:
For x > 2, f(x) = 5x + 2

Right hand limit at x = 2 is

= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 + f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 + (5x + 2) = 5*2 + 2 = 12,

For x < 2, f(x) = 7x – 2

Left hand limit at x = 2 is

= x$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 – f(x) = x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 – (7x – 2) = 7*2 – 2 = 12.

= 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)

So, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 f(x) = 12.

 

d.

Solution:
For x ≥ 2, f(x) = 3x + 1

Right hand limit at x = 2 is

= 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 = 7,

For x < 2, f(x) = 2x² – 1

Left hand limit at x = 2 is

= 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 = 7.

= 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)

So, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 f(x) = 7.

 

e.

Solution:
For x ≥ 1, f(x) = 2x + 1

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 + (2x + 1) = 2*1 + 1 = 3.

For x < 1, f(x) = 4x² – 1.

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 – (4x² – 1) = 4*1 – 1 = 3.

= 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)

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

 

f.

Solution:
For x ≥ 1, f(x) = 3x – 2

Right hand limit at x = 2 is

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 – 2) = 3*2 – 2 = 4,

For x < 2, f(x) = 2x² + 1

Left hand limit at x = 2 is

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.

= 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)

So, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 f(x) does not exist.

 

2.

a.

Solution:

LHL = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 f(2 – h)

= h$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 |2 – h – 2| = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 |h| = 0

RHL = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 f(2 + h)

= h$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 |2 + h – 2| = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 |h| = 0.

LHL = RHL

So, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 2 |x – 2| = 0.

 

b.

Solution:

RHL = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 f(0 + h)

= h$\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\left| {0 + {\rm{h}}} \right|}}{{0 + {\rm{h}}}}$ = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\rm{h}}}{{\rm{h}}}$ = 1.

LHL = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\left| {0 - {\rm{h}}} \right|}}{{0 - {\rm{h}}}}$ = h $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\rm{h}}}{{ - {\rm{h}}}}$ = –1.

LHL ≠ RHL

So, x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\left| {\rm{x}} \right|}}{{\rm{x}}}$ does not exist.