lim _{x rightarrow infty} xleft(log left(1+fr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) We are given the limit: $\lim _{x \rightarrow \infty} x\left(\log \left(1+\frac{x}{2}\right)-\log \frac{x}{2}\right)$.Using the property $\log a -

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) We are given the limit: $\lim _{x \rightarrow \infty} x\left(\log \left(1+\frac{x}{2}\right)-\log \frac{x}{2}\right)$.Using the property $\log a - \log b = \log \left(\frac{a}{b}\right)$,we have:$\lim _{x \rightarrow \infty} x \log \left(\frac{1+\frac{x}{2}}{\frac{x}{2}}\right) = \lim _{x \rightarrow \infty} x \log \left(\frac{2}{x} + 1\right)$.Let $t = \frac{1}{x}$. As $x \rightarrow \infty$,$t \rightarrow 0$. The expression becomes:$\lim _{t \rightarrow 0} \frac{\log (1+2t)}{t}$.Using the standard limit $\lim _{u \rightarrow 0} \frac{\log (1+u)}{u} = 1$,we multiply and divide by $2$:$\lim _{t \rightarrow 0} 2 \cdot \frac{\log (1+2t)}{2t} = 2 \cdot 1 = 2$.

$\lim _{x \rightarrow \infty} x\left(\log \left(1+\frac{x}{2}\right)-\log \frac{x}{2}\right) = $

Similar Questions

If $f(x) = \begin{cases} \frac{\sin(1+[x])}{[x]}, & \text{for } [x] \neq 0 \\ 0, & \text{for } [x] = 0 \end{cases}$ where $[x]$ denotes the greatest integer function,then $\lim_{x \rightarrow 0^{-}} f(x)$ is equal to

$\mathop {\lim }\limits_{x \to \infty } \left[ {\sqrt {x + \sqrt {x + \sqrt x } } - \sqrt x } \right]$ is equal to

If $\lim _{n \rightarrow \infty} x_n$ exists and is finite,$x_1=2$,$x_{n+1}=\frac{a+b x_n}{b+c x_n}$ for all $n \in N$,and $c > b > a > 0$,then $\lim _{n \rightarrow \infty} x_n =$

The value of $\mathop {\lim }\limits_{n \to \infty } \frac{1 - n^2}{\sum n}$ is

$\lim _{x \rightarrow 0} \frac{a^x-1}{\sin (x)} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module