# Engineering Mathematics Questions and Answers – Indeterminate Forms – 2

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) -∞

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) 32
b) 0
c) 43
d) –43

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) ba = 2
b) ab = 2
c) a = b
d) a = -b

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

Explanation: Form here 00
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

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.
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6. Find $$lt_{p\rightarrow\infty}\frac{p^5.p!}{5.6…(5+p)}$$
a) 4!
b) 5!
c) 0
d) ∞

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

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) 1cosh(1012345)
b) 90987
c) 1012345
d) ∞

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

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

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) 12
c) 0
d) 3132

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) √π2
d) 0

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) ∞

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

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