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one year ago

```							Dear student apply log on both sides y =x[(x/x+1)^x – 1/e] log y = logx + xlog(x/x+1) – lodx/e log y = xlog x/x+1 + loge log y = x log x/x+1 + loge log y =log e y = e Hope this helps
```
one year ago
```							hello madhav obviously vikas answer is wrong as he has applied logarithm rules incorrectly.this limit is actually a difficult limit. first, you need to write y= (x/(x+1))^x as e^logy= e^[x*log(x/(x+1))]. now, you need to write log(x/(x+1)) as log(1 – 1/(x+1)) and then do its series expansion using the std expansion log(1 – z)= – (z+z^2/2+z^3/3+......). obviously in our case z= 1/(x+1). then you need to multiply and divide by e and use the result lim p tends to 0 (e^p – 1)/p= 1. the limit will then become elementary.answer is 1/(2e).kindly approve :=)
```
one year ago
```							 apply log on both sides y =x[(x/x+1)^x – 1/e] log y = logx + xlog(x/x+1) – lodx/e log y = xlog x/x+1 + loge log y = x log x/x+1 + loge log y =log e y = e
```
9 months ago
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