# Engineering Mathematics Questions and Answers – Limits and Derivatives of Several Variables – 3

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This set of Engineering Mathematics online test focuses on “Limits and Derivatives of Several Variables – 3”.

1. limx → 1⁡ (x-1)Tan(πx2) is?
a) 0
b) –1π
c) –2π
d) 2π

Explanation:
$$\lim_{x\rightarrow 1}\frac{(x-1)sin(\frac{\pi x}{2})}{cos(\frac{\pi x}{2})}=\frac{0}{0}$$ (Indeterminate)
By L’Hospital rule
$$\lim_{x\rightarrow 1}\frac{(x-1)cos(\frac{\pi x}{2})\frac{\pi}{2}+sin(\frac{\pi x}{2})}{\frac{\pi}{2}sin(\frac{x\pi}{2})}=-\frac{2}{\pi}$$

2. Value of limit always be in the range of function.
a) True
b) False

Explanation: Because the range of f(x) = {x} is [0,1) and it value at limx → 1⁡ – f(x) is 1 which is not in its range.

3. Which of the following is a necessary Conditions of Sandwich rule?
a) All function must have common domain
b) All function must have common range
c) All function must have common domain and range both
d) Function must not have common domain and range

Explanation: Statement of sandwich theorem is, If Functions f(x), g(x) and h(x)
1. have Common Domain,
2. and, satisfy f(x) ≤ g(x) ≤ h(x) ∀ x ∈ D
Then if f(x) = h(x) = L
=> g(x) = L.

4. The value of limx → 0⁡⁡ [x]Cos(x), [x] denotes the greatest integer function _______
a) lies between 0 and 1
b) lies between -1 and 0
c) lies between 0 and 2
d) lies between -2 and 0

Explanation: limx → 0⁡⁡ [x]Cos(x)
We know that,
x-1 < [x] < x
Multiplying by Cos(x), we get
(x-1)Cos(x) < [x]Cos(x) < xCos(x)
Taking limits, we get
limx → 0 [(x-1)Cos(x)] < limx → 0 [x]Cos(x) < limx → 0[xCos(x)] => -1 < limx → 0 [x]Cos(x) < 0.

5. Value of limx → 0[(1+xex)/(1 – Cos(x))].
a) e
b) 1
c) 2
d) Can not be solved

Explanation: =>limx → 0[(1+xex)/(1 – Cos(x))] = 10 (Indeterminate)
=> By L’Hospital rule
=> limx → 0[(1+xex) / (Sin(x))] = 10 (Again indeterminate)
=> By L’ Hospital rule
=> limx → 0[((2+x)ex)/ (Cos(x))] = 2.

6. The value of $$\lim_{x\rightarrow 1}[x]cos(\frac{\pi(1-x)}{2})e^{1/(1-x)}$$, [x] denotes the greatest integer function.
a) 0
b) 1
c) ∞
d) -∞

Explanation:
$$\lim_{x\rightarrow 1}[x]cos(\frac{\pi(1-x)}{2})e^{1/(1-x)}$$
We know that
x-1 ≤ [x] ≤ x
Multiplying by Remaining term of question
$$(x-1)e^{1/(1-x)}cos(\frac{\pi(1-x)}{2})≤e^{1/(1-x)}cos(\frac{\pi(1-x)}{2})≤[x]≤xe^{1/(1-x)}cos(\frac{\pi(1-x)}{2})$$
$$\lim_{x\rightarrow 1}(x-1)e^{1/(1-x)}cos(\frac{\pi(1-x)}{2})≤\lim_{x\rightarrow 1}e^{1/(1-x)}cos(\frac{\pi(1-x)}{2})[x]$$
$$≤\lim_{x\rightarrow 1}xe^{1/(1-x)}cos(\frac{\pi(1-x)}{2})$$
By rearranging the terms of e1/(1-x) to e-1/(1-x)
$$\lim_{x\rightarrow 1}e^{-1/(x-1)}cos(\frac{\pi(1-x)}{2})x-1≤\lim_{x\rightarrow 1}e^{1/(1-x)}cos(\frac{\pi(1-x)}{2})[x]$$
$$≤\lim_{x\rightarrow 1}e^{-1/(x-1)}cos(\frac{\pi(1-x)}{2})x$$
$$0≤e^{-1/(x-1)}cos(\frac{\pi(1-x)}{2})[x]≤0$$
Hence by sandwich rule
$$\lim_{x\rightarrow 1}e^{1/(1-x)}cos(\frac{\pi(1-x)}{2})[x]=0$$

7. Evaluate limx → 0(1+Tan(x))Cot(x)
a) 1
b) e
c) ln(2)
d) e2

Explanation:
limx → 0(1+Tan(x))Cot(x) = limtan(x) → 0 (1+Tan(x))1Tan(x) = limt → 0 (1 + t)1t = e.

8. Evaluate limx → 1[(-xx + 1) / (xlog(x))].
a) ee
b) e
c) -1
d) e2

Explanation:
$$\lim{x\rightarrow 1}[(-x^x+1)/(xlog(x))]=(0/0)$$
By L’Hospital rule,
$$-\lim_{x\rightarrow 1}[x^x(1+xlog(x))/(1+xlog(x))]=-\lim_{x\rightarrow 1}[x^x]=-1$$

9. Find domain of n for which limx → 0enxCot(nx), has non zero value.
a) n ∈ (0,∞) ∩ (1,5)
b) n ∈ (-∞,∞) ∩ (1,5)
c) n ∈ (-∞,∞)
d) n ∈ (-∞,∞) ~ 5

Explanation:
$$\lim_{x\rightarrow 1}\frac{e^{nx}cos(nx)}{sin(nx)}=(1/0)$$
By L’hospital Rule we get
$$\Rightarrow \lim_{x\rightarrow 0}\frac{ne^{nx}(-sin(nx)+cos(nx))}{ncos(nx)}=n/n=1$$
Hence domain of n is n ∈ (-∞,∞).

10. Value of $$\frac{dSin(x)Cos(x)}{dx}$$ is
a) Cos(2x)
b) Sin(2x)
c) Cos2(2x)
d) Sin2(2x)

Explanation: $$\frac{dSin(x)Cos(x)}{dx} = Cos(x) \frac{dSin(x)}{dx} + Sin(x) \frac{dCos(x)}{dx}$$ = Cos2(x) – Sin2(x) = Cos(2x).

11. Evaluate $$\lim_{x\rightarrow\infty}(sin(\frac{1}{x})+cos(\frac{1}{x}))^x$$
a) 1
b) e
c) 0
d) e2

Explanation:
$$\lim_{x\rightarrow\infty}(sin(\frac{1}{x})+cos(\frac{1}{x}))^x$$
Putting x=1/y,
$$\Rightarrow \lim_{y\rightarrow 0}(sin(y)+cos(y))^{\frac{1}{y}}$$
$$\Rightarrow \lim_{y\rightarrow 0}((y-\frac{y^3}{3!}+\frac{y^5}{5!}-…)+(1-\frac{y^2}{2!}+\frac{y^4}{4!}-….))^{\frac{1}{y}}$$
Neglecting higher powers of y,(as y is limits to 0 which is very small hence higher power terms can be neglected)
$$\Rightarrow\lim_{y\rightarrow 0}(1+y)^{\frac{1}{y}}$$
=>e

12. If $$\lim_{x\rightarrow 0}\frac{(x(1+acos(x))-bsin(x))}{x^3}=1$$, then find the value of a and b.
a) 2.5, -1.5
b) -2.5, -1.5
c) -2.5, 1.5
d) 2.5, 1.5

Explanation:
$$\lim_{x\rightarrow 0}\frac{(x(1+acos(x))-bsin(x))}{x^3}=1$$
Expanding terms of cos(x) and sin(x) and rearranging we get,
$$\lim_{x\rightarrow 0}\frac{(1+a-b)x+(\frac{b}{6}-\frac{a}{2})x^3+(\frac{a}{24}-\frac{b}{120})x^5+….}{x^3}=1$$
Since, given limit is finite, hence coefficients of powers of x should be zero and x3 should be 1
⇒ 1 + a – b=0
b6a2 = 1
⇒ Solving the above two equations we get, a = -2.5, b = -1.5.

13. $$\lim_{x\rightarrow 0}\frac{ax^3+b sin(x)+c cos(x)}{x^5}=1$$, then find the value of a, b and c.
a) 1.37, -4.13, 4.13
b) 1.37, 4.13, -4.13
c) -1.37, 4.13, 4.13
d) 1.37, 4.13, 4.13

Explanation:
$$\lim_{x\rightarrow 0}\frac{ax^3+b sin(x)+c cos(x)}{x^5}=1$$
Now expanding the terms of sin(x) and cos(x) and rearranging in powers of x,x3 and x5 and so on,we get
=>$$\lim_{x\rightarrow 0}\frac{x(b+c)-x^3(\frac{b}{6}+\frac{c}{2}-a)+x^5(\frac{b}{120}+\frac{c}{24})+…}{x^5}$$
Now, coefficient of x and x3 should be zero and that of x5 should be 1, then
⇒ B + c = 0
b6 + c2 = a
b120 + c24 = 1
⇒ By solving these 3 equations, a = 1.37, b = 4.13, c = -4.13.

Sanfoundry Global Education & Learning Series – Engineering Mathematics.

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