Math, asked by kiara9514, 9 months ago

Hello guys

don't post irreverent answer

i⃟t⃟s⃟ k⃟i⃟a⃟r⃟a⃟ h⃟e⃟r⃟e⃟

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Answered by saounksh

✪ᴀɴsᴡᴇʀ✪

$\red{\star \star \star \blue{\boxed{\lim \limits_{x\to 0}{\left(\frac{3{x}^{2}+2}{7{x}^{2}+2}\right) }^{\frac{1}{{x}^{2}}} = \frac{1}{{e}^{2}}}}\star \star \star }$

᯽ғᴏʀᴍᴜʟᴀ᯽

$\green{\star \star \star \red{\boxed{\lim \limits_{x \to 0} {(x+1)}^{\frac{1}{x}} = e}}\star \star \star }$

★ᴇxᴘʟᴀɪɴᴀᴛɪᴏɴ★

$\lim \limits_{x\to 0}{\left(\frac{3{x}^{2}+2}{7{x}^{2}+2}\right)}^{\frac{1}{{x}^{2}}}$

$=\lim \limits_{x\to 0}{\left(\frac{\frac{3}{2}{x}^{2}+1}{\frac{7}{2}{x}^{2}+1}\right) }^{\frac{1}{{x}^{2}}}$

$=\lim \limits_{x\to 0}\frac{{\left(\frac{3}{2}{x}^{2}+1\right)}^{\frac{1}{{x}^{2} }}}{{\left(\frac{7}{2}{x}^{2}+1\right)}^{\frac{1}{{x}^{2}}}}$

$=\lim \limits_{x\to 0}\frac{{\left[{\left(\frac{3}{2}{x}^{2}+1\right)}^{\frac{1}{\frac{3}{2}{x}^{2}}}\right]}^{\frac{3}{2}}} {{\left[{\left(\frac{7}{2}{x}^{2}+1\right)}^{\frac{1}{\frac{7}{2}{x}^{2}} }\right]}^{\frac{7}{2}}}$

$=\frac{\lim \limits_{x\to 0}{\left[{\left(\frac{3}{2}{x}^{2}+1\right)}^{\frac{1}{\frac{3}{2}{x}^{2}} }\right]}^{\frac{3}{2}}}{\lim \limits_{x\to 0}{\left[{\left(\frac{7}{2}{x}^{2}+1\right)}^{\frac{1}{\frac{7}{2}{x}^{2}}} \right]}^{\frac{7}{2}}}$

$=\frac{{\left[\lim \limits_{\frac{3 }{2 }{x}^{2 }\to 0} {\left(\frac{3}{2}{x}^{2}+1\right)}^{\frac{1}{\frac{3}{2}{x}^{2}} }\right]}^{\frac{3}{2}}}{{\left[\lim \limits_{\frac{7}{2}{x}^{2 }\to 0}{\left(\frac{7}{2}{x}^{2}+1\right)}^{\frac{1}{\frac{7}{2}{x}^{2}}}\right]}^{\frac{7}{2}}}$

Put

$\frac{3}{2}{x}^{2} = y$

$\frac{7}{2}{x}^{2} = z$

$=\frac{{\left[\lim \limits_{y \to 0} {(y+1)}^{\frac{1}{y}}\right]}^{\frac{3}{2}}}{{\left[\lim \limits_{z \to 0}{(z + 1)}^{\frac{1}{z}}\right]}^{\frac{7}{2}}}$