This set of Engineering Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Indeterminate Forms – 2”.
1. Find \(lt_{x\rightarrow -2}\frac{sin(\frac{1+(\frac{(x+2)^2(x^2+1)}{x^3+3})}{x+2})}{(x+2)}\)
a) ∞
b) 0
c) 2
d) -∞
View Answer
Explanation: First evaluate
\(=lt_{x\rightarrow -2}\frac{ln(1+\frac{(x+2)^2(x^2+1)}{x^3+3})}{x+2}\)
\(=lt_{x\rightarrow -2}(\frac{1}{x+2})\times(\frac{(x+2)^2(x^2+1)}{(x^3+3)}-\frac{(x+2)^3(x^3+1)^2}{2.(x^3+3)^2}+…\infty)\)
\(=lt_{x\rightarrow -2}\times(\frac{(x+2)^2(x^2+1)}{(x^3+3)}-\frac{(x+2)^3(x^3+1)^2}{2.(x^3+3)^2}+…\infty)\)
We hence the form for the totle limit as
\(lt_{x\rightarrow a}\frac{sin(f(x))}{g(x)}\)=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 \(lt_{x\rightarrow 0}\frac{(3e^x-2e^{2x}-e^{3x})}{(e^x+e^{2x}-2e^{3x})}\)
a) 3⁄2
b) 0
c) 4⁄3
d) –4⁄3
View Answer
Explanation: Form is 0 / 0
Applying L hospitals rule we have
\(=lt_{x\rightarrow 0}\frac{3e^x-4e^{2x}-3e^{3x}}{e^x+2e^{2x}-6e^{3x}}\)
\(=\frac{3-4-3}{1+2-6}\)
\(=\frac{4}{3}\)
3. Find relation between a and b such that the following limit is got after a single application of L hospitals Rule \(lt_{x\rightarrow 0}\frac{ae^x+be^{2x}}{be^x+ae^{2x}}\)
a) b⁄a = 2
b) a⁄b = 2
c) a = b
d) a = -b
View Answer
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 \(lt_{x\rightarrow 0}\frac{2cos(2x)+3cos(5x)-5cos(19x)}{cos(4x)-cos(3x)}\)
a) -76
b) -6
c) -7
d) 0
View Answer
Explanation: Form here 0⁄0
Applying L hospitals rule we have
\(=lt_{x\rightarrow 0}\frac{4cos(2x)+15cos(5x)-95cos(19x)}{4cos(4x)-3cos(3x)}\)
\(\frac{4+15-95}{4-3}\) = -76.
5. Find how many rounds of differentiation are required to have finite limit for \(lt_{x\rightarrow 0}\frac{cos(ax)+cos(bx)-2cos(cx)}{cos(ax)+2cos(bx)-3cos(cx)}\) given that a ≠ b ≠ c
a) 3
b) 0
c) 2
d) 4
View Answer
Explanation: Applying L hospitals rule
\(=lt_{x\rightarrow 0}\frac{a.cos(ax)+b.cos(bx)-2c.cos(cx)}{a.cos(ax)+2b.cos(bx)-3c.cos(cx)}=lt_{x\rightarrow 0}\frac{a+b-2c}{a+2b-3c}\)
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 \(lt_{p\rightarrow\infty}\frac{p^5.p!}{5.6…(5+p)}\)
a) 4!
b) 5!
c) 0
d) ∞
View Answer
Explanation: \(=lt_{p\rightarrow\infty}\frac{1.2.3.4}{1.2.3.4}\times \frac{p^5.p!}{5.6…(5+p)}\)
\(=lt_{p\rightarrow\infty}\frac{4!.p^5.p!}{(p+5)!}\)
\(=lt_{p\rightarrow\infty}\frac{4!.p^5}{(p+5)(p+4)(p+3)(p+2)(p+1)}\)
\(=lt_{p\rightarrow\infty}(4!)\times\frac{p^5}{(p+5)(p+4)(p+3)(p+2)(p+1)}=4!\times(1)\)
= 4!
7. Find \(=lt_{x\rightarrow 0}\frac{sin(x)}{tan(x)}\)
a) 0
b) 1
c) ∞
d) 2
View Answer
Explanation: \(=lt_{x\rightarrow 0}\frac{\frac{sin(x)}{x}}{\frac{tan(x)}{x}}\)
\(=\frac{lt_{x\rightarrow 0}\frac{sin(x)}{x}}{lt_{x\rightarrow 0}\frac{tan(x)}{x}}\)
=1/1=1
8. Find \(lt_{x\rightarrow 1012345}(\frac{[sinh(x)]^2-[cosh(x)]^2}{[sinh(x)]^2+[cosh(x)]^2})\)
a) 1⁄cosh(1012345)
b) 90987
c) 1012345
d) ∞
View Answer
Explanation: \(lt_{x\rightarrow 1012345}\left ( \frac{1}{(\frac{(e^x+e^{-2x})^2+(e^x-e^{-x})^2}{4})}\right )\)
\(=lt_{x\rightarrow 1012345}\frac{1}{\frac{(e^{2x}+e^{-2x})}{2}}=\frac{1}{cosh(1012345)}\)
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
View Answer
Explanation: Drawing the graph of y = x and y = sin(x) we can write the limit value as 0.
10. Find \(lt_{x\rightarrow 0}\frac{sin(x^2)}{x}\)
a) ∞
b) -1
c) 0
d) 22
View Answer
Explanation: Expand into Taylor Series
\(=lt_{x\rightarrow 0}(\frac{1}{x})\times(\frac{x^2}{1!}-\frac{x^6}{3!}+..\infty)\)
\(=lt_{x\rightarrow 0}\times(\frac{x}{1!}-\frac{x^5}{3!}+..\infty)\)
=0
11. Find \(lt_{x\rightarrow -33}\frac{ln(x^3+68x^2+1222x+2179)-ln(x+1)}{(x^2+66x+1089)}\)
a) -33
b) 1⁄2
c) 0
d) 31⁄32
View Answer
Explanation: \(=lt_{x\rightarrow -33}\frac{ln(1+\frac{(x+33)^2(x+2)}{(x+1)})}{(x+33)^2}\)
\(=lt_{x\rightarrow -33}(\frac{1}{(x+33)^2})\times(\frac{(x+33)^2(x+2)}{(x+1)}-\frac{(x+33)^4(x+2)^2}{2(x+1)^2}….\infty)\)
\(=lt_{x\rightarrow -33}(\frac{(x+2)}{(x+1)}-\frac{(x+33)^2(x+2)^2}{2(x+1)^2}….\infty)\)
\(=lt_{x\rightarrow -33}(\frac{(x+2)}{(x+1)})=\frac{(-33+2)}{(-33+1)}\)
12. Find \(lt_{p\rightarrow\infty}\frac{p^{\frac{1}{2}}.p!}{\frac{1}{2}.\frac{3}{2}…(p+\frac{1}{2})}\)
a) √π
b) ∞
c) √π⁄2
d) 0
View Answer
Explanation: Using the Gauss definition of the Gamma function we have
\(\tau(x)=lt_{p\rightarrow\infty}\frac{p^x.p!}{x.(x+1)…(x+p)}\)
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 \(lt_{n\rightarrow\infty}(1+\frac{1}{n})^n\)
a) e
b) e – 1
c) 0
d) ∞
View Answer
Explanation: \(lt_{n\rightarrow\infty}(1+\frac{1}{n})^n=e^{lt_{n\rightarrow\infty}\frac{n}{n}}\)
\(lt_{n\rightarrow\infty}\frac{n}{n}\) = 1
= e
Sanfoundry Global Education & Learning Series – Engineering Mathematics.
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