Engineering Mathematics Multiple Choice Questions on “Indeterminate Forms – 1”.
1. Find ltx → ∞((1+frac{1}{x^2+2x+1})^{x^2+3x+1})
a) e
b) 1
c) e2
d) 1⁄e
Answer: a
Explanation: Use the form
lt x→ ∞(1 + f(x))g(x) = elt x→ ∞ f(x) * g(x)
Provided as x → ∞ we must have
f(x) → 0
g(x) → ∞
These conditions are met in our question
L = ltx → ∞((1+frac{1}{x^2+2x+1})^{x^2+3x+1} = e^{frac{x^2+3x+1}{x^2+2x+1}})
ltx → 0(frac{x^2+3x+1}{x^2+2x+1}=1)
L = e1 = e.
2. Find lt x → ∞((frac{ln(1+frac{(x+3)^3(2x+9)}{(4x^3+3)})}{x^3+3x^2+9x+27}))
a) 0
b) 1
c) Undefined
d) – 1⁄35
Answer: d
Explanation: The form here is of 0⁄0
Applying L hospitals rule would be really tough to differentiate. Hence we use the concept of Taylor Series
We know that ln(1+x)=(x-frac{x^2}{2}+frac{x^3}{3}-…infty)
Thus, we have
(=lt_{xrightarrow -3}(frac{1}{x^3+3x^2+9x+27}times (frac{(x+3)^3(2x+9)}{4x^3+3}-frac{(x+3)^6(2x+9)^2}{2(4x^3+3)^2}+..infty)))
(=lt_{xrightarrow -3}(frac{1}{x+3}^3times(frac{(x+3)^3(2x+9)}{4x^3+3}-frac{(x+3)^6(2x+9)^2}{2(4x^3+3)^2}+..infty)))
(=lt_{xrightarrow -3}(frac{(2x+9)}{4x^3+3}-frac{(x+3)^3(2x+9)^2}{2(4x^3+3)^2}+…infty))
All the terms except the first one go to zero, we now have
(=lt_{xrightarrow -3}frac{(2x+9)}{(4x^3+3)}=frac{2(-3)+9}{4(-3)^3+3}=frac{3}{-105})
(=-frac{1}{35})
3. Find ltn → ∞(sum_{a=0}^{n-1}frac{sin(frac{a}{n})}{n})
a) 1⁄a
b) 1
c) 1 – cos(1)
d) 0
Answer: c
Explanation: We use the concept of limit of a sum which is
(int_a^b f(x)dx=lt_{nrightarrow infty}(frac{b-a}{n})times(f(a)+f(a+frac{b-a}{n})….+f(a+frac{(n-1)(b-a)}{n})))
Thus we have
(int_0^1 sin(x)dx=lt_{nrightarrow infty}(frac{1}{n})times(f(0)+f(frac{1}{n})+….+f(frac{n-1}{n})))
(=lt_{nrightarrowinfty}frac{1}{n} times (sin(frac{1}{n})+sin(frac{2}{n})+…+sin(frac{n-1}{n})))
(int_0^1 sin(x)dx=lt_{nrightarrow infty}sum_{a=0}^{n-1}frac{sin(frac{a}{n})}{n})
It is enough to evaluate the integral
(int_{0}^{1} sin(x)dx=[-cos(x)]_0^1)=(cos(0)-cos(1))
=(1-cos(1))
4. Find ltx → 0((frac{ln(1+x^4)}{x}))
a) 1
b) -1
c) 0
d) Undefined
Answer: c
Explanation:
ltx → 0((frac{ln(1+x^4)}{x}) = lt_{xrightarrow 0} (frac{1}{x}) times (frac{x^4}{1}-frac{x^8}{2}+frac{x^{12}}{3}-…infty))
ltx → 0((frac{x^3}{1}-frac{x^7}{2}+frac{x^{11}}{3}-…infty))
= 0.
5. Find (lt_{xrightarrow 0}(frac{1}{sin^2(x)}))
a) 2
b) 1
c) 0
d) Undefined
Answer: d
Explanation:
=(lt_{xrightarrow 0}(frac{1}{(frac{1-cos(2x)}{2})})=lt_{xrightarrow 0}(frac{2}{1-cos(2x)}))
=((frac{2}{1-cos(0)})=frac{2}{0} rightarrow infty)
6. Find (lt_{xrightarrow infty}((frac{x^3+x^2+x}{x^3+x+1})^{x+3}))
a) e
b) e-1
c) 0
d) 1
Answer: a
Explanation:
=(lt_{xrightarrowinfty}(1+frac{x^2-1}{x^3+x+1^{x+3}})^{x+3}=e^{lt_{xrightarrowinfty}(frac{(x^2-1)(x+3)}{x^3+x+1})})
(lt_{xrightarrowinfty}(frac{(x^2-1)(x+3)}{x^3+x+1})=1)
= e1 = e.
7. Find (lt_{xrightarrow -2}(frac{sin((x-2)^2)}{(x+2)^2}))
a) 1
b) 0
c) ∞
d) 0⁄0
Answer: a
Explanation: We have 0⁄0 form
Now we have the form
(lt_{xrightarrow a}frac{sin(f(x))}{g(x)})=1
where f(x) → 0:g(x) → 0 as x → a
∴(lt_{xrightarrow -2}(frac{sin((x-2)^2)}{(x-2)^2})=1)
8. Find (lt_{nrightarrowinfty}sum_{a=1}^{n-1}(frac{ln(1+frac{a}{n})}{n}))
a) ln(2)
b) ln(4)
c) 3ln(2)
d) 1⁄a
Answer: b
Explanation: Using limit of sum we have
=(lt_{nrightarrowinfty}(frac{1}{n})times(ln(1+0)+ln(1+frac{1}{n})+ln(1+frac{2}{n})+…+ln(1+frac{(n-1)}{n})))
=(int_0^1 ln(1+x)dr=[(x+1)(ln(x+1)-1)]_0^1)
=(2ln(2)-1)-(ln(1)-1)=2ln(2)
=ln(4)
9. Find (lt_{nrightarrowinfty}frac{(1^a+2^a+…+(n-1)^a)}{n^{a+1}})
a) 1
b) 1⁄a + 1
c) 0
d) Undefined
Answer: b
Explanation: (int_0^1x^a dx=lt_{nrightarrowinfty}frac{1}{n}times((frac{1}{n})^a+(frac{2}{n})^a+…+(frac{n-1}{n})^a))
=(lt_{nrightarrowinfty}frac{(1^a+2^a+…+(n-1)^a)}{n^{a+1}})
It is enough to evaluatethis integral
(int_0^1 x^adx=[frac{x^{a+1}}{a+1}]_0^1)
=(frac{1}{a+1})
10. Find (lt_{xrightarrow -101}(frac{ln(x^2+20x+(x+101)^2(x^2+3))-ln(x^2+20x)}{x^2+202x+10201})).
a) 0
b) 1
c) ∞
d) (=frac{10204}{12221})
Answer: d
Explanation: (=lt_{xrightarrow -101}(frac{ln(1+frac{(x+101)^2(x^2+3)}{(x^2+20x)}}{(x+101)^2}))
Now expanding into Taylor series we have
(=lt_{xrightarrow -101}(frac{1}{(x+101)^2})times(frac{(x+101)^2(x^2+3)}{(x^2+20x)}-frac{(x+101)^4(x^2+3)^2}{2(x^2+20x)^2}+…infty))
(=lt_{xrightarrow -101}(frac{(x^2+3)}{(x^2+20x)}-frac{(x+101)^2(x^2+3)^2}{2(x^2+20x)^2}+….infty))
All others exceptthe first term tend to zero. Thus, we have
(=lt_{xrightarrow -101}(frac{(-101)^2+3}{(101)^2+20(101)}))
(=frac{10204}{12221})
11. Find (lt_{xrightarrow -5}(frac{tan^{-1}(x^2+6x+5)}{x^2+15x+50}))
a) 0
b) 1
c) -4⁄5
d) -1
Answer: c
Explanation: (=lt_{xrightarrow -5}frac{tan^{-1}((x+5)(x+1))}{(x+5)(x+10)})
Now expand into Taylor Series for tan-1(x)
(=lt_{xrightarrow -5}frac{1}{(x+5)(x+10)}times((x+5)(x+1)-frac{(x+5)^3(x+2)^3}{3}+frac{(x+5)^5(x+2)^5}{5}-..infty))
(=lt_{xrightarrow -5}(frac{(x+1)}{(x+10)}-frac{(x+5)^2(x+2)^3}{3(x+10)}+frac{(x+5)^4(x+2)^5}{5(x+10)}-…infty))
(=lt_{xrightarrow -5}frac{(-5)+1}{(-5)+10})
=(-frac{4}{5})
12. Find (lt_{xrightarrow 0}frac{sin((4x^3)tan^{-1}(x))}{x^4})
a) 1
b) 2
c) 4
d) 3
Answer: c
Explanation: Expand into Mclaurin series
(=lt_{xrightarrow 0}(frac{1}{x^4})times(frac{4x^3 tan^{-1}(x)}{1!}-frac{16x^6 tan^{-1}(x)}{3!}+…infty))
(=lt_{xrightarrow 0}(frac{4 tan^{-1}(x)}{x1!}-frac{16x^2 tan^{-1}(x)}{3!}+…infty))
Neglecting higher order terms (which go to zero) we have
(=lt_{xrightarrow 0}(frac{4 tan^{-1}(x)}{x!})=lt_{xrightarrow 0}(frac{4}{x})times(frac{x}{1}-frac{x^3}{3}+..infty))
(=lt_{xrightarrow 0}(frac{4 tan^{-1}(x)}{x!})=lt_{xrightarrow 0}(frac{4}{1}-frac{4x^2}{3}+…infty)=4)
13. Find (lt_{xrightarrow 0}frac{sin(sin(x))}{x})
a) 1
b) ∞
c) 0
d) -1
Answer: a
Explanation: (=lt_{xrightarrow 0}(frac{1}{x})times(frac{sin(x)}{1!}-frac{(sin(x))^3}{3!}+…infty))
(=lt_{xrightarrow 0}(frac{sin(x)}{x.1!}-frac{(sin(x))^3}{x.3!}+…infty))
(=lt_{xrightarrow 0}(frac{sin(x)}{x}-lt_{xrightarrow 0}frac{sin(x)}{x.3!}times(sin(x))^2+…infty))
(=lt_{xrightarrow 0}frac{sin(x)}{x}=1)
14. Find (=lt_{xrightarrow 0}frac{(a_nx^n+a_{n-1}x^{n-1}+…+a_1x+a_0)}{(b_nx^n+b_{n-1}x^{n-1}+…+b_1x+b_0)})
a) an ⁄ bn
b) ∞
c) No general form
d) bn ⁄ an
Answer: a
Explanation: (=lt_{xrightarrow 0}(frac{x^n}{x^n})timesfrac{(a_n+a_{n-1}frac{1}{x}+…+a_1frac{1}{x^{n-1}}+a_0frac{1}{x^{n}})}{(b_n+b_{n-1}frac{1}{x}+…+b_1frac{1}{x^{n-1}}+b_0frac{1}{x^{n}})})
(=frac{a_n}{b_n})