If f(x) = frac{1-x+sqrt{9x^2+10x+1}}{2x},then | vedclass

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

Question

(B) Given $f(x) = \frac{1-x+\sqrt{9x^2+10x+1}}{2x}$.To find $\lim_{x \rightarrow -1^{-}} f(x)$,let $x = -1-h$,where $h \rightarrow 0^{+}$ as $x \right

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(B) Given $f(x) = \frac{1-x+\sqrt{9x^2+10x+1}}{2x}$.To find $\lim_{x \rightarrow -1^{-}} f(x)$,let $x = -1-h$,where $h \rightarrow 0^{+}$ as $x \rightarrow -1^{-}$.Substituting $x = -1-h$ into the expression:$f(-1-h) = \frac{1-(-1-h) + \sqrt{9(-1-h)^2 + 10(-1-h) + 1}}{2(-1-h)}$$= \frac{2+h + \sqrt{9(1+2h+h^2) - 10 - 10h + 1}}{-2(1+h)}$$= \frac{2+h + \sqrt{9+18h+9h^2 - 10 - 10h + 1}}{-2(1+h)}$$= \frac{2+h + \sqrt{9h^2 + 8h}}{-2(1+h)}$Taking the limit as $h \rightarrow 0^{+}$:$\lim_{h \rightarrow 0^{+}} \frac{2+h + \sqrt{9h^2 + 8h}}{-2(1+h)} = \frac{2+0 + \sqrt{0}}{-2(1+0)} = \frac{2}{-2} = -1$.

If $f(x) = \frac{1-x+\sqrt{9x^2+10x+1}}{2x}$,then $\lim_{x \rightarrow -1^{-}} f(x) = $

Similar Questions

$\lim_{x \rightarrow -2^{+}} ([x]^2 - [x] - 2) + \lim_{x \rightarrow -3^{-}} ([x]^2 - 4[x] + 3) =$

$\lim _{x \rightarrow \infty}\left(\frac{x+6}{x+1}\right)^{x+4}$ is equal to

Let $L(m)$ be the $x$-coordinate of the left endpoint of the intersection of the graphs of $y = x^2 - 6$ and $y = m$,where $-6 < m < 6$. The value of $\mathop {\lim }\limits_{m \to 0} \left( {\frac{{L\left( { - m} \right) - L\left( m \right)}}{m}} \right)$ equals

$\lim _{n \rightarrow \infty}\left[\frac{1^3}{1-n^4}+\frac{2^3}{1-n^4}+\ldots +\frac{n^3}{1-n^4}\right]=$

The $\lim _{x \rightarrow \infty}\left(\frac{3 x-1}{3 x+1}\right)^{4 x}$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module