If mathop {lim }limits_{x to infty } left{ {l | vedclass

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

Question

(A) Given $\mathop {\lim }\limits_{x 	o \infty } \left\{ {\ln \left( {{x^2} + 5x} ight) - 2\ln \left( {cx + 1} ight)} ight\} = -2$. Using the p

Vedclass · Accepted Answer

$c = e$

Vedclass · Answer

(A) Given $\mathop {\lim }\limits_{x 	o \infty } \left\{ {\ln \left( {{x^2} + 5x} ight) - 2\ln \left( {cx + 1} ight)} ight\} = -2$. Using the property $\ln(a) - \ln(b) = \ln(a/b)$,we get: $\mathop {\lim }\limits_{x 	o \infty } \ln \left( \frac{x^2 + 5x}{(cx + 1)^2} ight) = -2$. Taking the exponential of both sides: $\mathop {\lim }\limits_{x 	o \infty } \frac{x^2 + 5x}{c^2x^2 + 2cx + 1} = e^{-2}$. Dividing the numerator and denominator by $x^2$ as $x 	o \infty$: $\frac{1}{c^2} = \frac{1}{e^2}$. Therefore,$c^2 = e^2$,which implies $c = e$ (since $cx+1$ must be positive for the logarithm to be defined as $x 	o \infty$,$c$ must be positive). Thus,$c = e$.

If $\mathop {\lim }\limits_{x \to \infty } \left\{ {\ln \left( {{x^2} + 5x} \right) - 2\ln \left( {cx + 1} \right)} \right\} = -2$,then:

Similar Questions

If $f(x) = \begin{cases} 4x-5, & x \leq 2 \\ x-k, & x > 2 \end{cases}$ then the value of $k$ for which $\lim_{x \rightarrow 2} f(x)$ exists is equal to:

$\lim _{x}$ ${\rightarrow -a} \frac{x^7+a^7}{x+a} = 7$ $\Rightarrow a = ?$

If $\lim _{x \rightarrow 5} \frac{x^{k}-5^{k}}{x-5}=500$,then the value of $k$,where $k \in N$ is

If $\lim _{x \rightarrow 0} \frac{2 a \sin x-\sin 2 x}{\tan ^{3} x}$ exists and is equal to $1$,then the value of $a$ is

If $\lim _{x \rightarrow 0} \frac{\cos (2 x)+a \cos (4 x)-b}{x^4}$ is finite,then $(a+b)$ is equal to :

Vedclass Test Series

Exam Paper Generator

Online Exam Module