# Engineering Mathematics Questions and Answers – Indeterminate Forms – 4

This set of Engineering Mathematics Multiple Choice Questions & Answers focuses on “Indeterminate Forms – 4”.

1. $$\lim_{x\rightarrow 0}⁡\frac{x^2 Sin(x) – e^{x^2}}{Cos⁡(x+π/2)}$$ is
a) 0
b) 1
c) 2
d) 3

Explanation:
$$\lim_{x\rightarrow 0}⁡\frac{x^2 Sin(x) – e^{x^2}}{Cos⁡(x+π/2)}$$=-1/0 (Indeterminate form)
By L’Hospital rule
$$\lim_{x\rightarrow 0}⁡\frac{x^2 Sin(x) – e^{x^2}}{Cos⁡(x+π/2)}$$ $$=\lim_{x\rightarrow 0}⁡\frac{x^2 Cos(x) + 2xSin(x) – 2xe^{x^2}}{-Sin(x+π/2)}$$= 0

2. Value of limx → 0⁡(1+Sin(x))Cosec(x)
a) e
b) 0
c) 1
d) ∞

Explanation: limx → 0⁡(1+Sin(x))Cosec(x)
Put sin(x) = t we get
limt → 0⁡(1+t)(1t)= e.

3. Value of limx → 0⁡(1+cot(x))sin(x)
a) e
b) e2
c) 1e
d) Can not be solved

Explanation:
$$\Rightarrow \lim_{x\rightarrow 0}(1+cot(x))^{sin(x)}=\lim_{x\rightarrow 0}(1+\frac{cos(x)}{sin(x)})^{sin(x)}$$
$$=\lim_{x\rightarrow 0}(1+\frac{cos(x)}{sin(x)})^{\frac{sin(x)}{cos(x)}cos(x)}$$
$$\Rightarrow \lim_{x\rightarrow 0}\left [(1+\frac{cos(x)}{sin(x)})^{\frac{sin(x}{cos(x)}}\right ]^{cos(x)}$$
$$\Rightarrow$$ Put cos(x)/sin(x)=t gives
$$\Rightarrow \lim_{t\rightarrow 0}\left [(1+t)^{\frac{1}{t}} \right ] ^{\lim_{x\rightarrow 0}cos(x)}$$
=>e1
=>e

4. $$\lim_{x\rightarrow\infty}f(x)^{g(x)}=\lim_{x\rightarrow\infty}f(x)^{\lim_{x\rightarrow\infty}g(x)}$$
a) True
b) False

Explanation: It is a property of limits.

5. $$ln(lim_{x\rightarrow\infty}\frac{f(x)}{g(x)})=\lim_{x\rightarrow\infty}ln(f(x))+\lim_{x\rightarrow\infty}ln(g(x))$$
a) True
b) False

Explanation:
$$ln(lim_{x\rightarrow\infty}\frac{f(x)}{g(x)})=ln(\frac{\lim_{x\rightarrow\infty}f(x)}{\lim_{x\rightarrow\infty}g(x)})$$
$$\Rightarrow ln(\lim_{x\rightarrow\infty}\frac{f(x)}{g(x)}) = \lim_{x\rightarrow\infty}ln(f(x)) – \lim_{x\rightarrow\infty}ln(g(x))$$
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6. Evaluate limx → 1⁡[(xx – 1) / (xlog(x))].
a) ee
b) e
c) 1
d) e2

Explanation: limx → 1⁡ [(xx – 1) / (xlog(x))] = (00)
By L hospital rule,
limx → 1⁡ [xx (1+xlog(x))/ (1+xlog(x))] = limx → 1⁡ [xx] = 1.

7. Find n for which $$\lim_{x\rightarrow 0}\frac{(cos(x)-1)(cos(x)-e^x)}{x^n}$$, has non zero value.
a) >=1
b) >=2
c) <=2
d) ~2

Explanation: $$\lim_{x\rightarrow 0}\frac{(cos(x)-1)(cos(x)-e^x)}{x^n}=(0/0)$$
By L’Hospital Rule two times we get
=>$$\lim_{x\rightarrow 0}\frac{sin(2x)+e^x(cos(x)+sin(x))}{n(n-1)x^{n-2}}$$
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

Explanation: y=$$\lim_{x\rightarrow 0}(sin(2x))^{tan^2(2x)}$$
Taking log of both side
$$ln y=\lim_{x\rightarrow 0}\frac{ln(sin(2x))}{cot^2(2x)}(0/0)$$
By L’Hospital Rule
$$ln y=-\lim_{x\rightarrow 0}\frac{2cos(2x)}{sin(2x).4.cosec^2(2x)cot(2x)}=-0.5\lim_{x\rightarrow 0}sin^2(2x)$$=-0.5
=>y=e-0.5

9. Evaluate $$\lim_{x\rightarrow\infty}\left [\frac{x-1}{x-2} \right ]^x$$.
a) 14
b) 13
c) 12
d) 1

Explanation: $$y=\lim_{x\rightarrow\infty}\left [\frac{x-1}{x-2} \right ]^x$$
$$ln y=\lim_{x\rightarrow\infty}xln\left [\frac{x-1}{x-2} \right ]$$
=>$$\lim_{x\rightarrow\infty}\frac{\left [\frac{x-1}{x-2} \right ]}{\frac{1}{x}}$$
By putting 1=1/y, we get
=>$$\lim_{y\rightarrow 0}\frac{ln\left [\frac{x-1}{x-2} \right ]}{y}=[ln(1/2)]/0$$ (i.e indeterminate)
Hence by applying L’Hospital rule
=>$$\lim_{y\rightarrow 0}\frac{ln\left [\frac{x-1}{x-2} \right ]}{y}=\lim_{y\rightarrow 0}\frac{\frac{2-y-1+y}{(2-y)^2}}{\frac{1-y}{2-y}}=\lim_{y\rightarrow 0}\frac{\frac{1}{2-y}}{\frac{1-y}{1}}=\lim_{y\rightarrow 0}(\frac{1}{(2-y)(1-y)})$$=1/2

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