Question

compute limit x tends to "-infinity" 2x²-x+3/x²-2x+5​

Answer 1

Answer:

JUST APPLY L. HOSPITAL RULE TILL YOU GET YOUR ANSWER IN CASE OF 0 BY 0 AND INFINITY♾️ BY INFINITY♾️

Answer 2

$\large\underline{\sf{Solution-}}$

Given expression is

$\purple{\rm :\longmapsto\:\displaystyle\lim_{x \to - \infty }\rm \frac{ {2x}^{2} - x + 3}{ {x}^{2} - 2x + 5}}$

To evaluate this limit, we use Method of Substitution

So, Substitute

$\purple{\rm :\longmapsto\:x = - y}$

$\purple{\rm :\longmapsto\:as \: x \: \to \: - \infty , \: so \: y \: \to \: \infty }$

So, above expression can be rewritten as

$\rm \:  =  \:\displaystyle\lim_{y \to \infty }\rm \frac{ {2( - y)}^{2} - ( - y)+ 3}{ {( - y)}^{2} - 2( - y)+ 5}$

$\rm \:  =  \:\displaystyle\lim_{y \to \infty }\rm \frac{2 {y}^{2} + y+ 3}{ {y}^{2} + 2y+ 5}$

$\rm \:  =  \:\displaystyle\lim_{y \to \infty }\rm \frac{ {y}^{2} \bigg[2+ \dfrac{1}{y} + \dfrac{3}{ {y}^{2}} \bigg]}{ {y}^{2} \bigg[1+ \dfrac{2}{y} + \dfrac{5}{ {y}^{2}} \bigg]}$

$\rm \:  =  \:\displaystyle\lim_{y \to \infty }\rm \frac{\bigg[2+ \dfrac{1}{y} + \dfrac{3}{ {y}^{2}} \bigg]}{\bigg[1+ \dfrac{2}{y} + \dfrac{5}{ {y}^{2}} \bigg]}$

$\rm \:  =  \: \dfrac{2 + 0 + 0}{1 + 0 + 0}$

$\rm \:  =  \: 2$

Hence,

$\purple{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to - \infty }\rm \frac{ {2x}^{2} - x + 3}{ {x}^{2} - 2x + 5}} = 2}}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

MORE TO KNOW

$\purple{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{sinx}{x} = 1}}}$

$\purple{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{tanx}{x} = 1}}}$

$\purple{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{log(1 + x)}{x} = 1}}}$

$\purple{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{ {e}^{x} - 1}{x} = 1}}}$

$\purple{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{ {a}^{x} - 1}{x} = loga}}}$