# Find the limit if it exists. If it does hot cxíst explain why Ih the case of ihf

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.
$\underset{x\to 1}{lim}\frac{\left({x}^{2}-1\right){e}^{\frac{1}{x-1}}}{\left(x-1\right)}$
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$\underset{x\to 1}{lim}\frac{\left({x}^{2}-1\right){e}^{\frac{1}{x-1}}}{\left(x-1\right)}$
$L.H.L=\underset{h\to 0}{lim}\frac{\left({\left(1-h\right)}^{2}-1\right){e}^{\frac{1}{x-h-x}}}{\left(x-h-x\right)}$
$=\underset{h\to 0}{lim}\frac{x+{h}^{2}-2h-x}{-h}{e}^{\frac{1}{h}}$
$=\underset{h\to 0}{lim}-\left(h-2\right){e}^{-\frac{1}{h}}$
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
Jeffrey Jordon