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  • Evaluate the limit: <img alt="\mathrm{\lim _{n \rightarrow 0} \frac{\sqrt{73} \cdot \log (1+n)}{\sqrt{1+n}-1}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim%20_%7Bn%20%5C

Evaluate the limit: \mathrm{\lim _{n \rightarrow 0} \frac{\sqrt{73} \cdot \log (1+n)}{\sqrt{1+n}-1}}

Option: 1

\sqrt{73}


Option: 2

\frac{2}{e}


Option: 3

2\sqrt{73}


Option: 4

\frac{4}{e}


Answers (1)

best_answer

\mathrm{\lim _{n \rightarrow 0} \frac{\sqrt{73} \cdot \log (1+n)}{\sqrt{1+n}-1}}
Rationalizing the denominator:

\mathrm{=\lim _{n \rightarrow 0} \frac{\sqrt{73} \cdot \log (1+n)}{(\sqrt{1+n}-1) \cdot \frac{(\sqrt{1+n}+1)}{(\sqrt{1+n}+1)}} \\}

\mathrm{=\lim _{n \rightarrow 0} \frac{\sqrt{73} \cdot \log (1+n)(\sqrt{1+n}+1)}{n} \\}

\mathrm{ \text { We know that: } \lim _{n \rightarrow 0} \frac{\log (1+n)}{n}=1 \text { so, } \\}

\mathrm{=1 \cdot \sqrt{73} \cdot(\sqrt{1}+1)}

\mathrm{=2\sqrt{73}}




 

Posted by

Gautam harsolia

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