1)\lim_{x\rightarrow \frac{\pi }{2}} sec^{-1}(sin x)
2)\lim_{x\rightarrow 0}cos\frac{1}{x}
3)\lim_{x\rightarrow 0}(1+sinx)^{1/x^{2}}
4) Let f(x+y2) = f(x)+f(y)2 for all real x and y.If f'(0) = -1 and f(0) = 1,find f(2)
5) If f(x+y3) = f(x)+f(y)+f(0)3 and f'(0) = 2, find f(x).
6)\lim_{x\rightarrow 1}(1-x^{2})log_{x^{2}}2
7)\lim_{x\rightarrow \frac{\pi }{2}}\frac{sinx - (sinx)^{sinx}}{1-sinx+ln(sinx)}
8) \lim_{x\rightarrow 0}\frac{\int_{-x}^{x}{f(t)dt}}{\int_{0}^{2x}{f(t+4)dt}}
9) \lim_{n\rightarrow \propto } \frac{a^{n}}{n!}
10)\lim_{x\rightarrow 0}\frac{sin\left\{x^{3} \right\}}{xsinx}
1@Adithya : how are you getting 0 to Q1??? pls explain....
solve this one too.......:-)
\lim_{n\rightarrow \infty }\left(1+\frac{1}{a_{1}} \right)\left(1+\frac{1}{a_{2}} \right)\left(1+\frac{1}{a_{3}} \right)...\left(1+\frac{1}{a_{n}} \right),
where a1 = 1 and an = n(1+an-1) for all n≥2
2624) and 5) - it is easy to notice that f(x)=ax+b
or use vivek's method
10) clearly LHL and RHL are unequal. hence limit does not exist .
1\lim_{n\rightarrow \infty }(11^{n}+5^{n}+6^{n})=\lim_{n\rightarrow \infty }11\left(1+\left(\frac{5}{11} \right)^{n}+\left(\frac{6}{11} \right)^{n} \right) = \lim_{n\rightarrow \infty }11.e^{0}=11
1708\hspace{-16}\mathbf{\left(1+\frac{1}{a_{1}}\right).\left(1+\frac{1}{a_{2}}\right).\left(1+\frac{1}{a_{3}}\right)........\left(1+\frac{1}{a_{n}}\right)=\prod_{r=1}^{n}\left(1+\frac{1}{a_{r}}\right)}$\\\\\\ $\mathbf{=\prod_{r=1}^{n}\left(\frac{a_{r}+1}{a_{r}}\right)=\prod_{r=1}^{n}\left(\frac{a_{r+1}}{a_{r}}\right).\left(\frac{1}{r+1}\right)}$\\\\\\ Using Recursive relation $\mathbf{a_{n}=n.(1+a_{n-1})}$\\\\ Put $\mathbf{n=r+1},$ We Get $\mathbf{a_{r+1}=(r+1).(1+a_{r})}$\\\\ $\mathbf{(1+a_{r})=\frac{a_{r+1}}{r+1}}$\\\\\\ So $\mathbf{\prod_{r=1}^{n}\left(\frac{a_{r+1}}{a_{r}}\right).\left(\frac{1}{r+1}\right)=\frac{a_{n+1}}{(n+1)!}}$ , bcz $\mathbf{a_{1}=1}$ (Given)\\\\\\ Now Let $\mathbf{b_{n+1}=\frac{a_{n+1}}{(n+1)!}}$\\\\\\ Then $\mathbf{b_{n}=\frac{a_{n}}{n!}}$\\\\\\ So $\mathbf{b_{n+1}-b_{n}=\frac{a_{n+1}}{(n+1)!}-\frac{a_{n}}{(n)!}=\frac{(n+1).(1+a_{n})}{(n+1)!}-\frac{a_{n}}{(n)!}=\frac{1}{(n)!}}$\\\\\\
\hspace{-16}$ So $\mathbf{b_{n+1}-b_{n}=\frac{1}{(n)!}}$\\\\\\ Put $\mathbf{n=1,2,3,4,5........}$ and Added..........$\\\\\\ \mathbf{\sum_{r=1}^{n}\left(b_{r+1}-b_{r}\right)=\sum_{r=1}^{n}\frac{1}{r!}}$\\\\\\ So $\mathbf{b_{n+1}-b_{1}=\sum_{r=1}^{n}\frac{1}{r!}}$\\\\\\ So $\mathbf{b_{n+1}=b_{1}+\sum_{r=1}^{n}\frac{1}{r!}=\sum_{r=0}^{n}\frac{1}{r!}}$\\\\\\ So $\mathbf{\lim_{n\rightarrow \infty} b_{n+1}=\lim_{n\rightarrow \infty}\frac{a_{n+1}}{(n+1)!}=\lim_{n\rightarrow \infty}\sum_{r=0}^{n}\frac{1}{r!}=e}$
If There is any Typing Mistake, Then plz Dont mind
Thanks
1i didn't understand.... wat's the derivative of e[x]?
pls post your complete solution...
2621) ex-1 < e[x] ≤ ex
use sandwich theorem .
(on diff. ex , it remains as it is, whereas exponent of x in xn decreases)
711. Set x = Î / 2 - a
As x → Π/2 + => a → 0 - ; x → Π/2 - => a → 0 +.
2. I think the limit does not exists because the values will oscillate between -1 and 1 as x → 0
3. It's 1∞ form
4. You can solve to find the functional equation via partial derivative method Or USE the property of the function in the definition of derivative and write f (a + h ) according as according to functional equation given aboove.
5. -Do-
6. Write logx2 2 = log 2 / log x2 , and then Use L - Hopital.
8. L - Hopital and Newton Leibnitz Lemma
9. The denominator grows more larger than numerator and hence the limit goes to 0.
10.USE Power Expansion Series.
2622) use L hospitals (including Newton Leibnitz in the numerator)
Ans. a=4
1in Q3, [.] denotes GIF
btw, i got 1 for the 3rd one ...but the answer given is -1....
how can it be -1?(since there is a modulus inside?)
1Part 2:
1)Find the value of\lim_{x\rightarrow infinity}\frac{x^{n}+nx^{n-1}+1}{e^{[x]}} where n is an integer
2)\lim_{x\rightarrow 0}\int_{0}^{x}{\frac{t^{2}dt}{(x-sinx)\sqrt{a+t}}}= 1 , then the value of a is
3)Let f_{p}(\alpha )=\left(cos\frac{\alpha }{p^{2}}+isin\frac{\alpha }{p^{2}} \right)\left(cos\frac{2\alpha }{p^{2}}+isin\frac{2\alpha }{p^{2}} \right)...\left(cos\frac{\alpha }{p}+isin\frac{\alpha }{p} \right) .
then \lim_{x\rightarrow infinity}\left[\left|f_{n}(2\pi ) \right| \right] =
PLS PROVIDE THE ANSWERS YOU GOT.....
71Yes Aditya You're right. But that is why the Sandwich has been used.
262i still dont think so.
that means you are supposing that n is an integer.
but then in its neighbourhood factorial will cease to exist.
1limn→c(n!) surely doesn't exisit. but what jciit has done is correct.
1Let \lambda \leq a<\lambda +1 where \lambda \in I^{+}
\Rightarrow \frac{a^{n}}{n!}=\frac{(a.a.a.a.....a)(a...a)}{(1.2.3.4...\lambda ).(\lambda +1)(\lambda +2)...n))}=\frac{a^{\lambda }}{\lambda !}.\frac{a}{\lambda +1}.\frac{a}{\lambda +2}...\frac{a}{n}
Clearly, \frac{a}{\lambda +1}>\frac{a}{\lambda +2}>\frac{a}{\lambda +3}>...\frac{a}{n}\Rightarrow \frac{a^{n}}{n!}<\frac{a^{\lambda }}{\lambda !}.\left(\frac{a}{\lambda +1} \right)^{n-\lambda }
Also, \frac{a}{\lambda +1}<1 \Rightarrow \lim_{n\rightarrow infinity}\left(\frac{a}{\lambda +1}\right)^{n-\lambda } = 0
Thus using sandwich theorem, the required limit = 0
2629. i think it should be does not exist, since n! exists only for natural nos.
