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lim_(x rarr+oo)-x+ln(1+xe^(x))=+oo

lim⁡x→+∞−x+ln⁡(1+xex)=+∞ \lim _{x \rightarrow+\infty}-x+\ln \left(1+x e^{x}\right)=+\infty

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Full solution

Q. lim⁡x→+∞−x+ln⁡(1+xex)=+∞ \lim _{x \rightarrow+\infty}-x+\ln \left(1+x e^{x}\right)=+\infty
  1. Simplify Inside Logarithm: Simplify the expression inside the logarithm:\newlineGiven −x+ln⁡(1+x⋅ex)-x + \ln(1 + x\cdot e^x), focus on the term inside the logarithm 1+x⋅ex1 + x\cdot e^x. As xx approaches infinity, exe^x grows much faster than xx, making x⋅exx\cdot e^x the dominant term. Thus, 1+x⋅ex1 + x\cdot e^x approximates to x⋅exx\cdot e^x.
  2. Substitute Simplified Expression: Substitute the simplified expression back: Replace 1+xex1 + x e^x with xexx e^x in the original limit expression, getting −x+ln⁡(xex)-x + \ln(x e^x).
  3. Apply Logarithmic Properties: Apply properties of logarithms:\newlineUsing the logarithmic identity ln⁡(ab)=ln⁡(a)+ln⁡(b)\ln(ab) = \ln(a) + \ln(b), rewrite ln⁡(xex)\ln(xe^x) as ln⁡(x)+ln⁡(ex)\ln(x) + \ln(e^x). Since ln⁡(ex)=x\ln(e^x) = x, the expression becomes −x+ln⁡(x)+x-x + \ln(x) + x.
  4. Simplify Final Expression: Simplify the expression:\newlineThe terms −x-x and xx cancel each other out, leaving ln⁡(x)\ln(x). As xx approaches infinity, ln⁡(x)\ln(x) also approaches infinity.

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