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

$\lim _{x \rightarrow 2}\left[\frac{1}{x-2}-\frac{2}{x^3-3 x^2+2 x}\right]$ is equal to

$\mathop {\lim }\limits_{x \to \infty } \frac{{(x - 1)(2x + 3)}}{{{x^2}}} = $

If $f(x) = \begin{cases} |x|+1, & x < 0 \\ 0, & x = 0 \\ |x|-1, & x > 0 \end{cases}$,for what value$(s)$ of $a$ does $\lim_{x \to a} f(x)$ exist?

$\mathop {\lim }\limits_{x \to \infty } \frac{{3{x^2} + 2x - 1}}{{2{x^2} - 3x - 3}} = $

If $l, m$ $(l < m)$ are roots of $ax^2 + bx + c = 0$,then $\lim_{x \rightarrow \alpha} \frac{|ax^2 + bx + c|}{ax^2 + bx + c} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module