Question

$\lim _{x \rightarrow 0} \frac{1}{x}\left(\sqrt{\frac{1}{x^{2}}+1}-\frac{1}{x}\right)+x \ln \left(1+a^{1 / x}\right), a>0, a \neq$ (A) a (B) (C) \(1+\mathrm{a}\) (D) None of these

Short answer

(D) None of these

Step-by-step solution

Separate the function into two parts

We will first find the limit of each part separately: 1. \(\lim_{x \rightarrow 0} \frac{1}{x}\left(\sqrt{\frac{1}{x^{2}}+1}-\frac{1}{x}\right)\) 2. \(\lim_{x \rightarrow 0} x \ln \left(1+a^{1 / x}\right)\)

Find the limit of the first part

To find the limit of the first part as x approaches 0, we can multiply and divide by the conjugate of the expression inside the bracket: \(\lim_{x \rightarrow 0} \frac{1}{x}\left(\sqrt{\frac{1}{x^{2}}+1}-\frac{1}{x}\right) \cdot \frac{\left(\sqrt{\frac{1}{x^{2}}+1}+\frac{1}{x}\right)}{\left(\sqrt{\frac{1}{x^{2}}+1}+\frac{1}{x}\right)}\) This simplifies to: \(\lim_{x \rightarrow 0} \frac{1}{x} \cdot \frac{(\frac{1}{x^2}+1)-\frac{1}{x^2}}{\sqrt{\frac{1}{x^2}+1}+\frac{1}{x}}\) Which further simplifies to: \(\lim_{x \rightarrow 0} \frac{1}{\sqrt{\frac{1}{x^2}+1}+\frac{1}{x}}\) We can now use the limit properties and substitution method: Limit of the first part = \(\frac{1}{\sqrt{\frac{1}{0^2}+1}+\frac{1}{0}} = \frac{1}{\sqrt{1}+\infty} = 0\)

Find the limit of the second part

To find the limit of the second part as x approaches 0, we can use the limit properties and substitution method: \(\lim_{x \rightarrow 0} x \ln \left(1+a^{1 / x}\right) = 0 \cdot \ln (1 + a^{\infty}) = 0 \cdot \ln (1 + a) = 0\)

Add the limits of both parts

Now we know the limits of both parts, we can add the limits to find the final answer: Final Limit = Limit of the first part + Limit of the second part = 0 + 0 = 0 The limit of this function as x approaches 0 is \(0\). So the correct answer is (D) None of these.