Engineering Mathematics Multiple Choice Questions on “Indeterminate Forms – 2”.
1. Find (lt_{xrightarrow -2}frac{sin(frac{1+(frac{(x+2)^2(x^2+1)}{x^3+3})}{x+2})}{(x+2)})
a) ∞
b) 0
c) 2
d) -∞
Answer: c
Explanation: First evaluate
(=lt_{xrightarrow -2}frac{ln(1+frac{(x+2)^2(x^2+1)}{x^3+3})}{x+2})
(=lt_{xrightarrow -2}(frac{1}{x+2})times(frac{(x+2)^2(x^2+1)}{(x^3+3)}-frac{(x+2)^3(x^3+1)^2}{2.(x^3+3)^2}+…infty))
(=lt_{xrightarrow -2}times(frac{(x+2)^2(x^2+1)}{(x^3+3)}-frac{(x+2)^3(x^3+1)^2}{2.(x^3+3)^2}+…infty))
We hence the form for the totle limit as
(lt_{xrightarrow a}frac{sin(f(x))}{g(x)})=1
Where f(x)→0 : g(x)→0 as x→a
This is true for the above problem
Thus, we can deduce the limit as
= 1
Hence, 2 is the right answer.
2. Find (lt_{xrightarrow 0}frac{(3e^x-2e^{2x}-e^{3x})}{(e^x+e^{2x}-2e^{3x})})
a) 3⁄2
b) 0
c) 4⁄3
d) –4⁄3
Answer: c
Explanation: Form is 0 / 0
Applying L hospitals rule we have
(=lt_{xrightarrow 0}frac{3e^x-4e^{2x}-3e^{3x}}{e^x+2e^{2x}-6e^{3x}})
(=frac{3-4-3}{1+2-6})
(=frac{4}{3})
3. Find relation between a and b such that the following limit is got after a single application of L hospitals Rule (lt_{xrightarrow 0}frac{ae^x+be^{2x}}{be^x+ae^{2x}})
a) b⁄a = 2
b) a⁄b = 2
c) a = b
d) a = -b
Answer: d
Explanation: Given differentiation is applied once we get
ae0 + be0 = 0 = a + b (numerator → zero)
be0 + ae0 = 0 = a + b (denominator → zero)
Thus the relation between (a, b) and is
a + b = 0
OR
a = -b.
4. Find (lt_{xrightarrow 0}frac{2cos(2x)+3cos(5x)-5cos(19x)}{cos(4x)-cos(3x)})
a) -76
b) -6
c) -7
d) 0
Answer: a
Explanation: Form here 0⁄0
Applying L hospitals rule we have
(=lt_{xrightarrow 0}frac{4cos(2x)+15cos(5x)-95cos(19x)}{4cos(4x)-3cos(3x)})
(frac{4+15-95}{4-3}) = -76.
5. Find how many rounds of differentiation are required to have finite limit for (lt_{xrightarrow 0}frac{cos(ax)+cos(bx)-2cos(cx)}{cos(ax)+2cos(bx)-3cos(cx)}) given that a ≠ b ≠ c
a) 3
b) 0
c) 2
d) 4
Answer: c
Explanation: Applying L hospitals rule
(=lt_{xrightarrow 0}frac{a.cos(ax)+b.cos(bx)-2c.cos(cx)}{a.cos(ax)+2b.cos(bx)-3c.cos(cx)}=lt_{xrightarrow 0}frac{a+b-2c}{a+2b-3c})
Assume now that
a + b + 2c = 0 and a + 2b – 3c = 0
We must have
a = c = b but given a ≠ c ≠ b
Thus, our assumption is false and a finite limit exists after first round of differentiation.
Hence, 2 is the right answer.
6. Find (lt_{prightarrowinfty}frac{p^5.p!}{5.6…(5+p)})
a) 4!
b) 5!
c) 0
d) ∞
Answer: a
Explanation: (=lt_{prightarrowinfty}frac{1.2.3.4}{1.2.3.4}times frac{p^5.p!}{5.6…(5+p)})
(=lt_{prightarrowinfty}frac{4!.p^5.p!}{(p+5)!})
(=lt_{prightarrowinfty}frac{4!.p^5}{(p+5)(p+4)(p+3)(p+2)(p+1)})
(=lt_{prightarrowinfty}(4!)timesfrac{p^5}{(p+5)(p+4)(p+3)(p+2)(p+1)}=4!times(1))
= 4!
7. Find (=lt_{xrightarrow 0}frac{sin(x)}{tan(x)})
a) 0
b) 1
c) ∞
d) 2
Answer: b
Explanation: (=lt_{xrightarrow 0}frac{frac{sin(x)}{x}}{frac{tan(x)}{x}})
(=frac{lt_{xrightarrow 0}frac{sin(x)}{x}}{lt_{xrightarrow 0}frac{tan(x)}{x}})
=1/1=1
8. Find (lt_{xrightarrow 1012345}(frac{[sinh(x)]^2-[cosh(x)]^2}{[sinh(x)]^2+[cosh(x)]^2}))
a) 1⁄cosh(1012345)
b) 90987
c) 1012345
d) ∞
Answer: a
Explanation: (lt_{xrightarrow 1012345}left ( frac{1}{(frac{(e^x+e^{-2x})^2+(e^x-e^{-x})^2}{4})}right ))
(=lt_{xrightarrow 1012345}frac{1}{frac{(e^{2x}+e^{-2x})}{2}}=frac{1}{cosh(1012345)})
9. Let f on (f(x)) denote the composition of f(x) with itself n number of times then the value of ltn → ∞ f on (sin(x)) =
a) -1
b) 2
c) ∞
d) 0
Answer: d
Explanation: Drawing the graph of y = x and y = sin(x) we can write the limit value as 0.
10. Find (lt_{xrightarrow 0}frac{sin(x^2)}{x})
a) ∞
b) -1
c) 0
d) 22
Answer: c
Explanation: Expand into Taylor Series
(=lt_{xrightarrow 0}(frac{1}{x})times(frac{x^2}{1!}-frac{x^6}{3!}+..infty))
(=lt_{xrightarrow 0}times(frac{x}{1!}-frac{x^5}{3!}+..infty))
=0
11. Find (lt_{xrightarrow -33}frac{ln(x^3+68x^2+1222x+2179)-ln(x+1)}{(x^2+66x+1089)})
a) -33
b) 1⁄2
c) 0
d) 31⁄32
Answer: d
Explanation: (=lt_{xrightarrow -33}frac{ln(1+frac{(x+33)^2(x+2)}{(x+1)})}{(x+33)^2})
(=lt_{xrightarrow -33}(frac{1}{(x+33)^2})times(frac{(x+33)^2(x+2)}{(x+1)}-frac{(x+33)^4(x+2)^2}{2(x+1)^2}….infty))
(=lt_{xrightarrow -33}(frac{(x+2)}{(x+1)}-frac{(x+33)^2(x+2)^2}{2(x+1)^2}….infty))
(=lt_{xrightarrow -33}(frac{(x+2)}{(x+1)})=frac{(-33+2)}{(-33+1)})
12. Find (lt_{prightarrowinfty}frac{p^{frac{1}{2}}.p!}{frac{1}{2}.frac{3}{2}…(p+frac{1}{2})})
a) √π
b) ∞
c) √π⁄2
d) 0
Answer: c
Explanation: Using the Gauss definition of the Gamma function we have
(tau(x)=lt_{prightarrowinfty}frac{p^x.p!}{x.(x+1)…(x+p)})
Where τ(x) is the Gamma function Using formula
(tau(x) times tau(1-x)=frac{pi.x}{sin(pi.x)})
put x=1/2 to get
((tau(frac{1}{2}))^2=frac{pi}{2.sin(frac{pi}{2})}=frac{pi}{2})
(tau(frac{1}{2})=sqrt{frac{pi}{2}})
13. Find (lt_{nrightarrowinfty}(1+frac{1}{n})^n)
a) e
b) e – 1
c) 0
d) ∞
Answer: a
Explanation: (lt_{nrightarrowinfty}(1+frac{1}{n})^n=e^{lt_{nrightarrowinfty}frac{n}{n}})
(lt_{nrightarrowinfty}frac{n}{n}) = 1
= e