## Rajasthan Board RBSE Class 11 Maths Chapter 10 Limits and Derivatives Ex 10.1

Question 1.

Show that left and right limits of function

at x = 1 are equal and their value is 1.

Solution:

From equation (i) and (ii)

L.H.L. = R.H.L.

⇒ f(1 – 0) = f(1 + 0) = 1

Hence, left and right limits of the given function at x = 1 are equal and their value is 1.

Hence Proved.

Question 2.

Is limit of the function at x = 0 ?

Solution:

It is clear from equation (i) and (ii),

L.H.L. ≠ R.H.L.

⇒ f(0 – 0) ≠ x(0 + 0)

Hence, limit of function does not exist at x = 0.

Question 3.

Prove that at x = 0, limits of function f(x) = | x | + | x – 1| exists.

Solution:

f(x) = | x | + | x – 1 |

Right limit at x – 0

R.H.L. = lim_{x→0+} + f(x) = f(0 + 0)

= lim_{h→0} f(0 + h)

= lim_{h→0 }|0 + h| + |0 + h – 1 | (h> 0)

= lim_{h→0 }| h | + | h – 1 |

= 0 + | 0 – 1 | = 1 …(i)

Left limit at x = 0

L.H.L. = lim_{h→0–} + f(x) = f(0 – 0)

= lim_{h→0} f(0 – h)

= lim_{h→0 }| 0 – h | + | 0 – h – 1 |

= lim_{h→0} | 0 – h | + | -(h – 1) |

= | -0 | + | -(0+ 1) |

= 0 + 1 = 1 …(ii)

It is clear from equation (i) and (ii),

L.H.L. = R.H.L.

⇒ f(0 – 0) = f(0 + 0) = 1

Hence, limit of function exists at x = 0. Hence Proved.

Question 4.

Prove that at x = 2, limits of function does not exists.

Solution:

Right limit at x = 2

R.H.L. = lim_{x→0} + f(x)

= f(2 + 0) = lim_{h→0} f(2 + h)

= lim_{h→0}[(2 + h)^{2} + (2 + h) + 1]

= lim_{h→0} [4 + h^{2} + 4h + 2 + h + 1]

= lim_{h→0} [h^{2} + 5h + 7]

= 0^{2} + 5(0) + 7 = 7 ….(i)

Left limit at x = 2

L.H.L. = lim_{x→0 – }f(x)

= f(2 – 0) = lim_{h→0} f(2 – h)

= lim_{h→0} [2 – h]

= 2 – 0 = 2 .

From equation (i) and (ii),

L.H.L. ≠ R.H.L.

⇒ f(2 – 0) ≠ f(2 + 0)

Hence, limit of function does not exist at x = 2.

Question 5.

Find the left and right limit of function f(x) = x cos ( 1/x) at x = 0.

Solution: