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) = x2 + 2x + 4 so that

Question 3.
Find
Solution:

Question 4.
Compute .
Solution:
For x ≠ 0, we know that -1 ≤ sin $$\frac{1}{x}$$ ≤ 1
∴ -x2 ≤ x2 . sin $$\frac{1}{x}$$ ≤ x2

Question 5.
Find .
Solution:
We define f : R → R by f(x) = x2 – 5 and g : R → R by g(x) = 4x + 10.

Question 6.
Find .
Solution:
Write F(x) = x3 – 6x2 + 9x
= x(x – 3)2 = (x – 3)
f(x) where f(x) = x(x – 3)
Write G(x) = x2 – 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) = x3 – 3x2 = x2(x – 3) = (x – 3)
f(x) where f(x) = x2,
and G(x) = x2 – 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 > $$\frac{1}{\sqrt{\varepsilon}}$$ ⇒ x2 > $$\frac{1}{\varepsilon}$$ ⇒ $$\frac{1}{x^{2}}$$ < ε

Question 22.
Show that
Solution:
Given k > 0, let ∞ = log k.
x > ∞ ⇒ ex ⇒ 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 an > 0, bm > 0, then show that f(x) = ∞ if n > m.
Solution:

Question 26.
Compute
Solution:
-1 ≤ sinx ≤ 1 ⇒ -1 ≤ -sinx ≤ 1
x2 – 1 ≤ x2 – sinx ≤ x2 + 1
Since x → ∞, suppose that the x2 – 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 x0 ∈ 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 x2, x ∈ R is continuous function.
Solution:
We define h : R → R by h(x) = x2 and
g : R → R by g(x) = cosx.
Now, for x ∈ R
have (goh)(x) = g(h(x)) = g(x2)
= cos x2 = 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).