Question

To solve without using l-hospital law.

Answer 1

Answer:

1

Step-by-step explanation:

Given

$\lim_{x \to 0}\;\;\;\frac{\log(2+x)-\log2}{\sqrt{1+x}-1}\\\;\\\lim_{x \to 0}\;\;\;\frac{(\log(2+x)-\log2)(\sqrt{1+x}+1)}{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}\\\;\\\lim_{x \to 0}\;\;\;\frac{(\log(2+x)-\log2)(\sqrt{1+x}+1)}{1+x-1}\\\;\\\lim_{x \to 0}\;\;\;\frac{(\log(2+x)-\log2)(\sqrt{1+x}+1)}{x}\\\;\\\lim_{x \to 0}\;\;\;\frac{(\log\frac{2+x}{2})(\sqrt{x+1}+1)}{x}\\\;\\\lim_{x \to 0}\;\;\;\frac{(\log(1+\frac{x}{2}))(\sqrt{x+1}+1)}{x}$

$\lim_{x \to 0}\;\;\;\frac{(\frac{x}{2}-\frac{x^2}{2.2}+..............)(\sqrt{x+1}+1)}{x}\\\;\\\lim_{x \to 0}\;\;\;(\frac{1}{2}-\frac{x}{2.2}+..............)(\sqrt{x+1}+1)\\\;\\=(\frac{1}{2}+0+0)(\sqrt{0+1}+1)\\\;\\=\frac{1}{2}\times2\\\;\\=1$

Note:-

$\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-............$

Answer 2

Given

$\begin{gathered}\lim_{x \to 0}\;\;\;\frac{\log(2+x)-\log2}{\sqrt{1+x}-1}\\\;\\\lim_{x \to 0}\;\;\;\frac{(\log(2+x)-\log2)(\sqrt{1+x}+1)}{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}\\\;\\\lim_{x \to 0}\;\;\;\frac{(\log(2+x)-\log2)(\sqrt{1+x}+1)}{1+x-1}\\\;\\\lim_{x \to 0}\;\;\;\frac{(\log(2+x)-\log2)(\sqrt{1+x}+1)}{x}\\\;\\\lim_{x \to 0}\;\;\;\frac{(\log\frac{2+x}{2})(\sqrt{x+1}+1)}{x}\\\;\\\lim_{x \to 0}\;\;\;\frac{(\log(1+\frac{x}{2}))(\sqrt{x+1}+1)}{x}\end{gathered}$

$\begin{gathered}\lim_{x \to 0}\;\;\;\frac{(\frac{x}{2}-\frac{x^2}{2.2}+..............)(\sqrt{x+1}+1)}{x}\\\;\\\lim_{x \to 0}\;\;\;(\frac{1}{2}-\frac{x}{2.2}+..............)(\sqrt{x+1}+1)\\\;\\=(\frac{1}{2}+0+0)(\sqrt{0+1}+1)\\\;\\=\frac{1}{2}\times2\\\;\\=1\end{gathered}$

Note:-

$\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-............log$