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124 Questions around this concept.
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The value of so that the fanction
becomes continuous is equal to
The value of $ \lim _{x \rightarrow 0} \frac{(1+x)^{1 / x}-e}{x} $ is
$
\lim _{n \rightarrow \infty} \frac{\left(1^2-1\right)(n-1)+\left(2^2-2\right)(n-2)+\ldots .+\left((n-1)^2-(n-1)\right) \cdot 1}{\left(1^3+2^3+\ldots . .+n^3\right)-\left(1^2+2^2+\ldots . .+n^2\right)}
$
is equal to:
If $\lim_{x\to 0}\frac{\ln \left ( 1+x \right )+\alpha \sin x+\frac{x^{2}}{2}}{x^{3}}= \beta \left ( \in R \right )$ ,
then the value of $\alpha +\beta$ is
Define a sequence $\left(a_n\right)$ by $a_1=5, a_n=a_1 a_2 \ldots \ldots a_{n-1}+4$ for $n>1$. Then $\lim _{n \rightarrow \infty} \frac{\sqrt{a_n}}{a_{n-1}}$
Define a sequence $\left\{a_n\right\}_{n \geq 0}$ by $a_n=\sqrt{\frac{1+a_{n-1}}{2}}$ for $n \geq 1, a_0=\cos \theta \neq 1$. Then $\lim _{n \rightarrow \infty} 4^n\left(1-a_n\right)$ equals
Limit Using Expansion (Part 1)
Using the expansions is one of the methods to find the limits. The following expansion formulas which are also known as Taylor series, are very useful in evaluating various limits.
(i) $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots \ldots$.
(ii) $a^x=1+x\left(\log _e a\right)+\frac{x^2}{2!}\left(\log _e a\right)^2+\frac{x^3}{3!}\left(\log _e a\right)^3+\ldots \ldots \ldots$
(iii) $\log (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots \ldots \ldots$.
(iv) $\log (1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-\ldots \ldots \ldots$.
(v) $(1+x)^n=1+n x+\frac{n(n-1)}{2!} x^2+\frac{n(n-1)(n-2)}{3!} x^3+\ldots \ldots \ldots$.
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