x tends to infinity সংক্রান্ত
The values of limn→∞n5+24−n2+13n4+25−n3+12\displaystyle\lim_{n\rightarrow \infty}\dfrac{\sqrt[4]{n^5+2}-\sqrt[3]{n^2+1}}{\sqrt[5]{n^4+2}-\sqrt[2]{n^3+1}}n→∞lim5n4+2−2n3+14n5+2−3n2+1 is?
111
000
−1-1−1
∞\infty∞
limn→∞n5+24−n2+13n4+25−n3+13=limn→∞n541+2/n5−n231+1/n2n4/51+2/n4−n3/21+1n3=limn→∞n23(n7/21+2n5−1+1n2)n→∞n1/5(1+2n4−n7101+1n3)n23=limn→∞n−215×limn→∞(n7/21+2n3−1+1n21+2n4−n7101+1/n3)=0×limn→∞n7121+2n5−1+1n21+2n4−n7161+1n3=0 \begin{array}{l}\lim _{n \rightarrow \infty} \frac{\sqrt[4]{n^{5}+2}-\sqrt[3]{n^{2}+1}}{\sqrt[5]{n^{4}+2}-\sqrt[3]{n^{3}+1}} \\ =\lim _{n \rightarrow \infty} \frac{n^{\frac{5}{4}} \sqrt{1+2 / n^{5}}-n^{\frac{2}{3}} \sqrt{1+1 / n^{2}}}{n^{4 / 5} \sqrt{1+2 / n^{4}}-n^{3 / 2} \sqrt{1+\frac{1}{n^{3}}}} \\ =\lim _{n \rightarrow \infty} n^{\frac{2}{3}}\left(n^{7 / 2} \sqrt{1+\frac{2}{n^{5}}}-\sqrt{1+\frac{1}{n^{2}}}\right) \\ n \rightarrow \infty \frac{n^{1 / 5}\left(\sqrt{1+\frac{2}{n^{4}}}-n^{\frac{7}{10}} \sqrt{1+\frac{1}{n^{3}}}\right)}{n^{\frac{2}{3}}} \\ =\lim _{n \rightarrow \infty} n^{-\frac{2}{15}} \times \lim _{n \rightarrow \infty}\left(\frac{n^{7 / 2} \sqrt{1+\frac{2}{n^{3}}}-\sqrt{1+\frac{1}{n^{2}}}}{\sqrt{1+\frac{2}{n^{4}}}-n^{\frac{7}{10}} \sqrt{1+1 / n^{3}}}\right) \\ =0 \times \lim _{n \rightarrow \infty} \frac{n^{\frac{7}{12}} \sqrt{1+\frac{2}{n^{5}}}-\sqrt{1+\frac{1}{n^{2}}}}{\sqrt{1+\frac{2}{n^{4}}}-n^{\frac{7}{16}} \sqrt{1+\frac{1}{n^{3}}}}=0 \\\end{array} limn→∞5n4+2−3n3+14n5+2−3n2+1=limn→∞n4/51+2/n4−n3/21+n31n451+2/n5−n321+1/n2=limn→∞n32(n7/21+n52−1+n21)n→∞n32n1/5(1+n42−n1071+n31)=limn→∞n−152×limn→∞(1+n42−n1071+1/n3n7/21+n32−1+n21)=0×limn→∞1+n42−n1671+n31n1271+n52−1+n21=0
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The value of limx→∞x(x{ln(x)−ln(x+1)}+1)\displaystyle\lim_{x\rightarrow \infty}x(x\{ln (x)-ln (x+1)\}+1)x→∞limx(x{ln(x)−ln(x+1)}+1) is?
If f(x)=13(f(x+1)+5f(x+2))f(x) = \dfrac {1}{3}\left (f (x + 1) + \dfrac {5}{f(x + 2)}\right )f(x)=31(f(x+1)+f(x+2)5) and f(x)>0f(x) > 0f(x)>0 for all xϵRx \epsilon RxϵR, then limx→∞f(x)\displaystyle \lim_{x\rightarrow \infty} f(x)x→∞limf(x) is
Let f:(π2,π2)→R,f(x)={limn→∞(tanx)2n+x2sin2x+(tanx)2n;x≠01;x=0,n∈N f:\left(\frac{\pi}{2}, \frac{\pi}{2}\right) \rightarrow R, f(x)=\left\{\begin{array}{cc}\lim _{n \rightarrow \infty} \frac{(\tan x)^{2 n}+x^{2}}{\sin ^{2} x+(\tan x)^{2 n}} ; & x \neq 0 \\ 1 ; & x=0\end{array}, n \in N\right. f:(2π,2π)→R,f(x)={limn→∞sin2x+(tanx)2n(tanx)2n+x2;1;x=0x=0,n∈N. Which of the following holds good?
If ϕ(x)=limn→∞x2nf(x)+g(x)1+x2n\phi (x) =\displaystyle \lim_{n \rightarrow \infty} \frac{x^{2n} f(x) + g(x)}{1 + x^{2n}}ϕ(x)=n→∞lim1+x2nx2nf(x)+g(x), then
