1. 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
2. Find \(lt_{x\rightarrow 1012345}(\frac{[sinh(x)]^2-[cosh(x)]^2}{[sinh(x)]^2+[cosh(x)]^2})\)
a) 1⁄cosh(1012345)
b) 90987
c) 1012345
d) ∞
Explanation: \(lt_{x\rightarrow 1012345}\left ( \frac{1}{(\frac{(e^x+e^{-2x})^2+(e^x-e^{-x})^2}{4})}\right )\)
\(=lt_{x\rightarrow 1012345}\frac{1}{\frac{(e^{2x}+e^{-2x})}{2}}=\frac{1}{cosh(1012345)}\)
3. 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.
4. 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
5. Find \(lt_{x\rightarrow -33}\frac{ln(x^3+68x^2+1222x+2179)-ln(x+1)}{(x^2+66x+1089)}\)
a) -33
b) 1⁄2
c) 0
d) 31⁄32
Explanation: \(=lt_{x\rightarrow -33}\frac{ln(1+\frac{(x+33)^2(x+2)}{(x+1)})}{(x+33)^2}\)
\(=lt_{x\rightarrow -33}(\frac{1}{(x+33)^2})\times(\frac{(x+33)^2(x+2)}{(x+1)}-\frac{(x+33)^4(x+2)^2}{2(x+1)^2}….\infty)\)
\(=lt_{x\rightarrow -33}(\frac{(x+2)}{(x+1)}-\frac{(x+33)^2(x+2)^2}{2(x+1)^2}….\infty)\)
\(=lt_{x\rightarrow -33}(\frac{(x+2)}{(x+1)})=\frac{(-33+2)}{(-33+1)}\)
6. 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) √π⁄2
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
\((\tau(\frac{1}{2}))^2=\frac{\pi}{2.sin(\frac{\pi}{2})}=\frac{\pi}{2}\)
\(\tau(\frac{1}{2})=\sqrt{\frac{\pi}{2}}\)
7. 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
8. \(\lim_{x\rightarrow 0}\frac{x^2 Sin(x) – e^{x^2}}{Cos(x+π/2)}\) 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
9. 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)(1⁄t)= e.
10. Value of limx → 0(1+cot(x))sin(x)
a) e
b) e2
c) 1⁄e
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