UPSC 2022 Maths Optional Paper 1 Q1c — Step-by-Step Solution

10 marks · Section A

Indeterminate forms · Calculus · asked 4× in 13 yrs · Read the full method →

Question

Evaluate $\displaystyle\lim_{x\to\infty}(e^x+x)^{1/x}$ .

Logarithm + factor out dominant term $e^x$ ; or L’Hôpital on $\ln(e^x+x)/x$ .

Form: $\infty^0$ as $x\to\infty$ (base grows, exponent $\to 0$ ). Take logarithm.

Let $L=\lim_{x\to\infty}(e^x+x)^{1/x}$ . Then

\ln L=\lim_{x\to\infty}\dfrac{\ln(e^x+x)}{x}.

Factor $e^x$ :

e^x+x=e^x\!\left(1+\dfrac{x}{e^x}\right).

$\ln(e^x+x)=x+\ln(1+x/e^x)$ .

As $x\to\infty$ , $x/e^x\to 0$ , so $\ln(1+x/e^x)\to\ln 1=0$ .

\dfrac{\ln(e^x+x)}{x}=\dfrac{x+\ln(1+x/e^x)}{x}=1+\dfrac{\ln(1+x/e^x)}{x}.

As $x\to\infty$ , the second term $\to 0/\infty=0$ .

So $\ln L=1$ , giving $L=e$ .

\boxed{\;\lim_{x\to\infty}(e^x+x)^{1/x}=e.\;}