# Engineering Mathematics Questions and Answers – Indeterminate Forms – 3

This set of Engineering Mathematics MCQs focuses on “Indeterminate Forms – 3”.

1. What are Intermediate Forms?
a) Forms(f(x)/g(x)) whose limits x tends to ‘a’ can give rational number directly
b) Forms(f(x)/g(x)) whose limits x tends to ‘a’ can give finite number directly
c) Forms(f(x)/g(x)) whose limits x tends to ‘a’ can not be infinite output or cannot be solved directly
d) Forms(f(x)/g(x)) whose limits x tends to ‘a’ can gives finite output

Explanation:
$$\lim_{x\rightarrow a}$$= 00 = 0 = 0 = are all intermediate forms.
Special Rules are made to solve these forms. They cannot be solved directly.

2. L’Hospital Rule states that
a) If $$\lim_{x\rightarrow a}\frac{f(x)}{g(x)}$$ is an indeterminate form than $$\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}$$ if $$\lim_{x\rightarrow a} \frac{f'(x)}{g'(x)}$$ has a finite value
b) $$\lim_{x\rightarrow a}\frac{f(x)}{g(x)}$$ always equals to $$\lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}$$
c) $$\lim_{x\rightarrow a}\frac{f(x)}{g(x)}$$ if an indeterminate form than cannot be solved
d) $$\lim_{x\rightarrow a}\frac{f(x)}{g(x)}$$ if an indeterminate form than it is equals to zero.

Explanation: According to L’Hospital Rule, if $$\lim_{x\rightarrow a}\frac{f(x)}{g(x)}$$ is indeterminate and $$\lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}$$ has a finite value then $$\lim_{x\rightarrow a}\frac{f(x)}{g(x)} = \lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}$$. It is helpful in solving limits of indeterminate forms.

3. If f(x) = x2 – 3x + 2 and g(x) = x3 – x2 + x – 1 than find value of $$\lim_{x\rightarrow 1}\frac{f(x)}{g(x)}$$?
a) 0.5
b) 1
c) -.5
d) -1

Explanation:
$$\lim_{x\rightarrow 1}\frac{f(x)}{g(x)}=\lim_{x\rightarrow 1}\frac{x^2-3x+2}{x^3-x^2+x-1}$$=(1-3+2)/(1-1+1-1)=0/0
Hence this gives indeterminent forms hence by L’Hospital Rule
$$\lim_{x\rightarrow 1}\frac{f(x)}{g(x)}=\lim_{x\rightarrow 1}\frac{f'(x)}{g'(x)}=\lim_{x\rightarrow 1}\frac{2x-3}{3x^2-2x+1}=\frac{2-3}{3-2+1}=-\frac{1}{2}$$
Hence
$$\lim_{x\rightarrow 1}\frac{f(x)}{g(x)}=-\frac{1}{2}$$

4. If f(x) = Tan(x)-1 and g(x) = Sin(x) – Cos(x) than find value of limx → π4 f(x)g(x)
a) -√2
b) √(-2)
c) √2
d) √3

Explanation:
$$\lim_{x\rightarrow \pi/4}\frac{f(x)}{g(x)}=\lim_{x\rightarrow \pi/4}\frac{tan(x)-1}{sin(x)-cos(x)}=\frac{(1-1)}{\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}}=0/0$$
Hence this gives indeterminate forms hence by L’Hospital Rule
$$\lim_{x\rightarrow \pi/4}\frac{f(x)}{g(x)}=\lim_{x\rightarrow \pi/4}\frac{f'(x)}{g'(x)}=\lim_{x\rightarrow \pi/4}\frac{sec^2}{cos(x)+sin(x)}=\sqrt{2}$$
Hence
$$\lim_{x\rightarrow \pi/4}\frac{f(x)}{g(x)}=\sqrt{2}$$

5. If f(x) = Tan(x) and g(x) = ex – 1 than find value of limx → 0⁡ f(x)g(x)
a) 1
b) 0
c) -1
d) 2

Explanation:
$$\lim_{x\rightarrow 0}\frac{f(x)}{g(x)}=\lim_{x\rightarrow 0}\frac{tan(x)}{e^x-1}=0/0$$ (Indeterminate Form)
By L’Hospital’s rule
$$\lim_{x\rightarrow 0}\frac{f(x)}{g(x)}=\lim_{x\rightarrow 0}\frac{f'(x)}{g'(x)}=\lim_{x\rightarrow 0}\frac{sec^2(x)}{e^x}$$=1

6. Find the value of $$\lim_{x\rightarrow \infty}(1+\frac{1}{x})^x$$
a) e-1
b) e
c) e + 1
d) 1

Explanation:
$$\lim_{x\rightarrow \infty}(1+\frac{1}{x})^x$$=1 (Indeterminate form)
Hence
Let t=1/x
->y=$$\lim_{x\rightarrow\infty}(1+\frac{1}{x})^x=\lim_{t\rightarrow 0}(1+t)^{1/t}$$
Taking log on both side,
ln(y) = $$\lim_{t\rightarrow 0} \frac{ln(1+t)}{t}$$ = 0/0 (Indeterminate form)
Applying L’Hospital Rule again
ln(y) = $$\lim_{t\rightarrow 0} \frac{1}{1+t}$$ = 1
=> y = $$\lim_{x\rightarrow\infty}(1+\frac{1}{x})^x$$ = e

7. If f(x) = x3 + 3x2 + Sin(x) and g(x) = ex – 1 than find value of $$\lim_{x\rightarrow 0}f(x)^{\frac{1}{g(x)}}$$
a) e6e
b) e(e/6)
c) e6
d) e(6/e)

Explanation:
y=$$lim_{x\rightarrow 0}f(x)^{(\frac{1}{g(x)})} = lim_{x\rightarrow 0}[x^3+3x^2+sin(x)]^{\frac{1}{e^x-1}}$$
Taking ln on both side
ln(y)=$$lim_{x\rightarrow 0}\frac{ln(x^3+3x^2+sin(x))}{e^x-1}$$=ln(0)/0 (Indeterminate Form)
By L’Hospital Rule
$$lim_{x\rightarrow 0}\frac{3x^+6x+cos(x)}{e^x(x^3+3x^2+sin(x))}=1/0$$
Again using L’Hospital rule
ln(y)=$$\lim_{x\rightarrow 0}\frac{6x+6+sin(x)}{e^x(x^3+3x^2+sin(x))+e^x(3x^2+6x+cos(x))}=6$$
y=e6

8. If f(x) = Sin(x) and g(x) = x than find value of limx → 0⁡ f(x)g(x)
a) -1
b) 0
c) 1
d) 2

Explanation: limx → 0⁡ f(x)g(x) = limx → 0⁡ sin(x)x = ⁡00 (Indeterminate forms)
By L’Hospital rule
limx → 0⁡ cos(x)1 = 1.

9. If f(x) = sin(x)cos(x) and g(x) = x2 than find value of limx → 0⁡ f(x)g(x)
a) 2
b) 0
c) -1
d) Cannot be found

Explanation:
$$\lim_{x\rightarrow 0}\frac{f(x)}{g(x)}=\lim_{x\rightarrow 0}\frac{sin(x)cos(x)}{x^2}$$ = 0/0
Henece, by L’Hospital rule
$$\lim_{x\rightarrow 0}\frac{f(x)}{g(x)}=\lim_{x\rightarrow 0}\frac{cos(2x)}{2x}=\frac{1}{0}$$
Again by L’Hospital rule
$$\lim_{x\rightarrow 0}\frac{f(x)}{g(x)}=\lim_{x\rightarrow 0}\frac{sin(2x)}{1}=0$$

10. If f(x) = ex + xcos(x) and g(x) = Sin(x) than find value of limx → 0⁡ f(x)g(x)
a) 2
b) 1
c) 3
d) 4

Explanation:
$$\lim_{x\rightarrow 0}\frac{f(x)}{g(x)}=\lim_{x\rightarrow 0}\frac{e^x+xcos(x)}{sin(x)}=\lim_{x\rightarrow 0}\frac{e^x}{sin(x)}+\lim_{x\rightarrow 0}\frac{xcos(x)}{sin(x)}$$=1/0+0/0, i.e both are indeterminate
i.e., applying L’Hospital Rule
$$\lim_{x\rightarrow 0}\frac{e^x}{sin(x)}=\lim_{x\rightarrow 0}\frac{e^x}{cos(x)}+\lim_{x\rightarrow 0}\frac{cos(x)-xsin(x)}{cos(x)}$$=1+1=2

11. Find the value of $$\lim_{x\rightarrow 0}\frac{xsin(x-1)}{(x-1)e^x}$$ is, where {x} is the fractional part of x.
a) 2e
b) 1e
c) 0
d) –1e

Explanation:
$$\lim_{{x}\rightarrow 1^+}\frac{{x}sin(x-1)}{(x-1)e^x}$$
$$=\lim_{x\rightarrow 1^+}{x}\lim_{x\rightarrow 1^+}{\frac{sin(x-1)}{x-1}} \lim_{x\rightarrow 1^+}\frac{1}{e^x}=\lim_{x\rightarrow 1^+}{x}1.\frac{1}{e}=\frac{1}{e}$$
because, $$\lim_{x\rightarrow 1^+}{x}=1$$.

Sanfoundry Global Education & Learning Series – Engineering Mathematics.

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