Students get through Maths 1B Important Questions Inter 1st Year Maths 1B Limits and Continuity Important Questions which are most likely to be asked in the exam.

## Intermediate 1st Year Maths 1B Limits and Continuity Important Questions

Question 1.

Evaluate

Solution:

Question 2.

Compute

Solution:

Write f(x) = \(\frac{x-2}{x^{3}-8}\) x ≠ 2 so that

f(x) = \(\frac{x-2}{x^{3}-8}\) = \(\frac{1}{x^{2}+2 x+4}\)

Write h(x) = x^{2} + 2x + 4 so that

Question 3.

Find

Solution:

Question 4.

Compute .

Solution:

For x ≠ 0, we know that -1 ≤ sin \(\frac{1}{x}\) ≤ 1

∴ -x^{2} ≤ x^{2} . sin \(\frac{1}{x}\) ≤ x^{2}

Question 5.

Find .

Solution:

We define f : R → R by f(x) = x^{2} – 5 and g : R → R by g(x) = 4x + 10.

Question 6.

Find .

Solution:

Write F(x) = x^{3} – 6x^{2} + 9x

= x(x – 3)^{2} = (x – 3)

f(x) where f(x) = x(x – 3)

Write G(x) = x^{2} – 9 = (x – 3) (x + 3)

= (x – 3) g(x) where g(x) = x + 3

∴ \(\frac{F(x)}{G(x)}=\frac{(x-3) f(x)}{(x-3) g(x)}=\frac{f(x)}{g(x)}\)

and g(3) = 6 ≠ 0.

If F and G are polynomials such that f(x) = (x – a)^{k}, G(x) = (x – a)^{k} g(x) for some k ∈ N and for some polynomials f(x) and g(x) with

Question 7.

Find .

Solution:

We write F(x) = x^{3} – 3x^{2} = x^{2}(x – 3) = (x – 3)

f(x) where f(x) = x^{2},

and G(x) = x^{2} – 5x + 6 = (x – 3)(x – 2)

= (x – 3) g(x) where g(x) = x – 2,

with g(3) = 3 – 2 = 1 ≠ 0.

∴ by applying Theorem g(a) ≠ 0

Question 8.

Show that and (x ≠ 0).

Solution:

Question 9.

Let f : R → R be defined by

f(x) = \(\begin{cases}2 x-1 & \text { if } x<3 \\ 5 & \text { if } x \geq 3\end{cases}\) show that .

Solution:

Question 10.

Show that .

Solution:

Observe that \(\sqrt{x^{2}-4}\) is not defined over (-2, 2)

Question 11.

Solution:

Question 12.

Show that .

Solution:

Question 13.

Find

Solution:

For 0 < |x| < 1, we have

Question 14.

Compute

Solution:

For 0 < |x| < 1

Question 15.

Show that

Solution:

Question 16.

Compute [Mar 13] (a > 0, b > 0, b ≠ 1).

Solution:

Question 17.

Compute , b ≠ 0, a ≠ b.

Solution:

Question 18.

Compute

Solution:

Question 19.

Compute

Solution:

Question 20.

Evaluate

Solution:

Question 21.

Show that

Solution:

Given ε > 0, choose ∞ = \(\frac{1}{\sqrt{\varepsilon}}\) > 0

x > ∞ ⇒ x > \(\frac{1}{\sqrt{\varepsilon}}\) ⇒ x^{2} > \(\frac{1}{\varepsilon}\) ⇒ \(\frac{1}{x^{2}}\) < ε

Question 22.

Show that

Solution:

Given k > 0, let ∞ = log k.

x > ∞ ⇒ e^{x} ⇒ e^{∞} = k

Question 23.

Compute

Solution:

Question 24.

Evaluate .

Solution:

Question 25.

If f(x) \(\frac{a_{n} x^{n}+\ldots+a_{1} x+a_{0}}{b_{m} x^{m}+\ldots+b_{1} x+b_{0}}\) when a_{n} > 0, b_{m} > 0, then show that f(x) = ∞ if n > m.

Solution:

Question 26.

Compute

Solution:

-1 ≤ sinx ≤ 1 ⇒ -1 ≤ -sinx ≤ 1

x^{2} – 1 ≤ x^{2} – sinx ≤ x^{2} + 1

Since x → ∞, suppose that the x^{2} – 2 > 0

Question 27.

Show that f(x) = [x] (x ∈ R is continuous at only those real numbers that are not integers.

Solution:

Case i) : If a ∈ z, f(a) = (a) = a

∴ f is not continuous at x = a ∈ z.

Case ii) : If a ∉ z, then ∃ n ∈ z such that n < a < n + 1 then f(a) = (a) = n.

Question 28.

If f : R → R is such that f(x + y) = f(x) + f(y) for all x, y ∈ R then f is continuous on R if it is continuous at a single point in R.

Solution:

Let f be continuous at x_{0} ∈ R

∴ f is continuous at x.

Since x ∈ R is arbitrary, f is continuous on R.

Question 29.

Check the continuity of the function f given below at 1 and 2.

f(x) = \(\left\{\begin{array}{cl}

x+1 & \text { if } x \leq 1 \\

2 x & \text { if } 1<x<2 \\

1+x^{2} & \text { if } x \geq 2

\end{array}\right.\)

Solution:

∴ f is continuous at x = 1

f is not continuous at x = 2.

Question 30.

Show that the function f defined on R by f(x) = Cos x^{2}, x ∈ R is continuous function.

Solution:

We define h : R → R by h(x) = x^{2} and

g : R → R by g(x) = cosx.

Now, for x ∈ R

have (goh)(x) = g(h(x)) = g(x^{2})

= cos x^{2} = f(x)

Since g and h continuous on their respective domains, by Theorem

Let A, B, ⊆ R.

Let f : A → R be continuous on A and let

g : B → R be continuous on B.

If f(A) ⊆ B then the composite function

gof : A → R is continous on A.

It follows that a continuous function on R.

Question 31.

Show that the function f defined on R by f(x) = |1 + 2x + |x||, x ∈ R is a continuous function.

Solution:

We define g : R → R by

g(x) = 1 + 2x + |x|, x ∈ R,

and h : R → R by h(x) = |x|, x ∈ R. Then

(hog) (x) = h(g(x)) = h(1 + 2x + |x|)

= |1 + 2x + |x|| = f(x).