If a 0 and lim _{x rightarrow a} frac{a^x - | vedclass

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

Question

(B) Given the limit: $\lim _{x \rightarrow a} \frac{a^x - x^a}{x^x - a^a} = -1$. Applying $L^{\prime}$Hospital's rule,we differentiate the numerator a

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) Given the limit: $\lim _{x \rightarrow a} \frac{a^x - x^a}{x^x - a^a} = -1$. Applying $L^{\prime}$Hospital's rule,we differentiate the numerator and denominator with respect to $x$: Numerator derivative: $\frac{d}{dx}(a^x - x^a) = a^x \ln a - a x^{a-1}$. Denominator derivative: $\frac{d}{dx}(x^x - a^a) = x^x(1 + \ln x)$. Substituting $x = a$: $\frac{a^a \ln a - a \cdot a^{a-1}}{a^a(1 + \ln a)} = -1$. $\frac{a^a \ln a - a^a}{a^a(1 + \ln a)} = -1$. Dividing numerator and denominator by $a^a$: $\frac{\ln a - 1}{1 + \ln a} = -1$. $\ln a - 1 = -(1 + \ln a)$. $\ln a - 1 = -1 - \ln a$. $2 \ln a = 0$. $\ln a = 0 \Rightarrow a = e^0 = 1$.

If $a > 0$ and $\lim _{x \rightarrow a} \frac{a^x - x^a}{x^x - a^a} = -1$,then $a$ is equal to

Similar Questions

Let $f(1) = g(1) = k$ and their $n^{th}$ derivatives $f^{(n)}(1), g^{(n)}(1)$ exist and are not equal for some $n$. If $\lim _{x \rightarrow 1} \frac{f(1) g(x) - f(1) - g(1) f(x) + g(1)}{g(x) - f(x)} = 4$,then the value of $k$ is:

$\mathop {\lim }\limits_{x \to 0} \frac{{\sin x - x}}{{{x^3}}} = $

Let $f(x)$ be a differentiable function and $f^{\prime}(4)=5$. Then, $\lim _{x \rightarrow 2} \frac{f(4) - f\left(x^{2}\right)}{x-2}$ equals

$\mathop {\lim }\limits_{\theta \to {0^ + }} {(\sin \theta )^{(\sin \theta - {{\sin }^2}\theta )}}$ is

Evaluate the limit: $\lim _{x \rightarrow 2} \frac{x^3-x^2-x-2}{2 x^3-3 x^2-3 x+2}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module