 # Find the limit if it exists. If it does hot cxíst explain why Ih the case of ihf arenceabigns 2021-10-19 Answered
Find the limit if it exists. If it does hot cxíst explain why Ih the case of ihfinite limits, malk suve that you check both the left and right limit at the point in the question.
$\underset{x\to 1}{lim}\frac{\left({x}^{2}-1\right){e}^{\frac{1}{x-1}}}{\left(x-1\right)}$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it gotovub
$\underset{x\to 1}{lim}\frac{\left({x}^{2}-1\right){e}^{\frac{1}{x-1}}}{\left(x-1\right)}$
$L.H.L=\underset{h\to 0}{lim}\frac{\left({\left(1-h\right)}^{2}-1\right){e}^{\frac{1}{x-h-x}}}{\left(x-h-x\right)}$
$=\underset{h\to 0}{lim}\frac{x+{h}^{2}-2h-x}{-h}{e}^{\frac{1}{h}}$
$=\underset{h\to 0}{lim}-\left(h-2\right){e}^{-\frac{1}{h}}$
Taking limit:
$L.H.L=0$
$R.H.L.=\underset{h\to 0}{lim}\frac{\left({\left(1+h\right)}^{2}-1\right){e}^{\frac{1}{x+h-x}}}{x+h-x}$
$=\underset{h\to 0}{lim}\frac{x+{h}^{2}+2h-x}{h}{e}^{\frac{1}{h}}$
$=\underset{h\to 0}{lim}\left(h+2\right){e}^{\frac{1}{h}}$
Taking limit
$R.H.L=\mathrm{\infty }$ (does not exist finity)
$⇒L.H.L\ne R.H.L$
$⇒\underset{x\to 1}{lim}\frac{\left({x}^{2}-1\right){e}^{\frac{1}{x-1}}}{\left(x-1\right)}$ does not exist
###### Not exactly what you’re looking for? Jeffrey Jordon