Engineering Mathematics Multiple Choice Questions & Answers focuses on “Indeterminate Forms – 4”.
1. (lim_{xrightarrow 0}frac{x^2 Sin(x) – e^{x^2}}{Cos(x+π/2)}) is
a) 0
b) 1
c) 2
d) 3
Answer: a
Explanation:
(lim_{xrightarrow 0}frac{x^2 Sin(x) – e^{x^2}}{Cos(x+π/2)})=-1/0 (Indeterminate form)
By L’Hospital rule
(lim_{xrightarrow 0}frac{x^2 Sin(x) – e^{x^2}}{Cos(x+π/2)})
(=lim_{xrightarrow 0}frac{x^2 Cos(x) + 2xSin(x) – 2xe^{x^2}}{-Sin(x+π/2)})= 0
2. Value of limx → 0(1+Sin(x))Cosec(x)
a) e
b) 0
c) 1
d) ∞
Answer: a
Explanation: limx → 0(1+Sin(x))Cosec(x)
Put sin(x) = t we get
limt → 0(1+t)(1⁄t)= e.
3. Value of limx → 0(1+cot(x))sin(x)
a) e
b) e2
c) 1⁄e
d) Can not be solved
Answer: a
Explanation:
(Rightarrow lim_{xrightarrow 0}(1+cot(x))^{sin(x)}=lim_{xrightarrow 0}(1+frac{cos(x)}{sin(x)})^{sin(x)})
(=lim_{xrightarrow 0}(1+frac{cos(x)}{sin(x)})^{frac{sin(x)}{cos(x)}cos(x)})
(Rightarrow lim_{xrightarrow 0}left [(1+frac{cos(x)}{sin(x)})^{frac{sin(x}{cos(x)}}right ]^{cos(x)} )
(Rightarrow) Put cos(x)/sin(x)=t gives
(Rightarrow lim_{trightarrow 0}left [(1+t)^{frac{1}{t}} right ] ^{lim_{xrightarrow 0}cos(x)})
=>e1
=>e
4. (lim_{xrightarrowinfty}f(x)^{g(x)}=lim_{xrightarrowinfty}f(x)^{lim_{xrightarrowinfty}g(x)})
a) True
b) False
Answer: a
Explanation: It is a property of limits.
5. (ln(lim_{xrightarrowinfty}frac{f(x)}{g(x)})=lim_{xrightarrowinfty}ln(f(x))+lim_{xrightarrowinfty}ln(g(x)))
a) True
b) False
Answer: b
Explanation:
(ln(lim_{xrightarrowinfty}frac{f(x)}{g(x)})=ln(frac{lim_{xrightarrowinfty}f(x)}{lim_{xrightarrowinfty}g(x)}))
(Rightarrow ln(lim_{xrightarrowinfty}frac{f(x)}{g(x)}) = lim_{xrightarrowinfty}ln(f(x)) – lim_{xrightarrowinfty}ln(g(x)))
6. Evaluate limx → 1[(xx – 1) / (xlog(x))].
a) ee
b) e
c) 1
d) e2
Answer: c
Explanation: limx → 1 [(xx – 1) / (xlog(x))] = (0⁄0)
By L hospital rule,
limx → 1 [xx (1+xlog(x))/ (1+xlog(x))] = limx → 1 [xx] = 1.
7. Find n for which (lim_{xrightarrow 0}frac{(cos(x)-1)(cos(x)-e^x)}{x^n}), has non zero value.
a) >=1
b) >=2
c) <=2
d) ~2
Answer:b
Explanation: (lim_{xrightarrow 0}frac{(cos(x)-1)(cos(x)-e^x)}{x^n}=(0/0))
By L’Hospital Rule two times we get
=>(lim_{xrightarrow 0}frac{sin(2x)+e^x(cos(x)+sin(x))}{n(n-1)x^{n-2}})
Hence, limit have non zero limit, if n ≠ 0 and (n-1) ≠ 0 and (n-2) >= 0 means n >= 2.
8. Find the value of limx → 0(Sin(2x))Tan2 (2x)?
a) e0.5
b) e-0.5
c) e-1
d) e
Answer: b
Explanation: y=(lim_{xrightarrow 0}(sin(2x))^{tan^2(2x)})
Taking log of both side
(ln y=lim_{xrightarrow 0}frac{ln(sin(2x))}{cot^2(2x)}(0/0))
By L’Hospital Rule
(ln y=-lim_{xrightarrow 0}frac{2cos(2x)}{sin(2x).4.cosec^2(2x)cot(2x)}=-0.5lim_{xrightarrow 0}sin^2(2x))=-0.5
=>y=e-0.5
9. Evaluate (lim_{xrightarrowinfty}left [frac{x-1}{x-2} right ]^x).
a) 1⁄4
b) 1⁄3
c) 1⁄2
d) 1
Answer: c
Explanation: (y=lim_{xrightarrowinfty}left [frac{x-1}{x-2} right ]^x)
(ln y=lim_{xrightarrowinfty}xlnleft [frac{x-1}{x-2} right ])
=>(lim_{xrightarrowinfty}frac{left [frac{x-1}{x-2} right ]}{frac{1}{x}})
By putting 1=1/y, we get
=>(lim_{yrightarrow 0}frac{lnleft [frac{x-1}{x-2} right ]}{y}=[ln(1/2)]/0) (i.e indeterminate)
Hence by applying L’Hospital rule
=>(lim_{yrightarrow 0}frac{lnleft [frac{x-1}{x-2} right ]}{y}=lim_{yrightarrow 0}frac{frac{2-y-1+y}{(2-y)^2}}{frac{1-y}{2-y}}=lim_{yrightarrow 0}frac{frac{1}{2-y}}{frac{1-y}{1}}=lim_{yrightarrow 0}(frac{1}{(2-y)(1-y)}))=1/2