My friend at school showed me some math competition problems, and a specific problem caught my eye:
Evaluate this limit: [MATH]\lim_{x\to 0} \left(\frac{\sqrt{1+\sin^2 x^2} - \cos^3 x^2}{x^3 \tan x}\right)[/MATH]
So, i decided to try and answer the problem. After a few hours, i've found a solution for it, but i feel a bit unsure. Here's my solution:
[MATH]\rightarrow[/MATH] Using L'Hopital's rule,
=[MATH]\lim_{x\to 0} \left(\frac{\frac{2x\sin x^2\cos x^2}{\sqrt{1+\sin^2 x^2}} + 6x\cos^2 x^2 \sin x^2}{3x^2 \tan x + x^3\sec^2 x}\right)[/MATH][MATH]\rightarrow[/MATH]Since [MATH]\sin 2x = 2\sin x \cos x[/MATH], Then [MATH]\sin 2x^2 = 2\sin x^2 \cos x^2[/MATH]
=[MATH]\lim_{x\to 0} \left(\frac{\frac{x\sin 2x^2}{\sqrt{1+\sin^2 x^2}} + 3x\sin 2x^2 \cos x^2 }{3x^2 \tan x + \frac{x^3}{\cos^2 x}}\right)[/MATH][MATH]\rightarrow[/MATH][MATH]\cos^2 x = 1 - \sin^2 x[/MATH]
=[MATH]\lim_{x\to 0} \left(\frac{\frac{x\sin 2x^2}{\sqrt{1+\sin^2 x^2}} + 3x\sin 2x^2 \cos x^2 }{3x^2 \tan x + \frac{x^3}{1 - \sin^2 x}}\right)[/MATH][MATH]\rightarrow[/MATH]Trigonometric limit properties,
=[MATH]\lim_{x\to 0} \left(\frac{\frac{x\sin 2x^2}{\sqrt{(1+\sin^2 x^2) . (\frac{x^4}{x^4})}}. (\frac{2x^2}{2x^2}) + 3x\sin 2x^2 \cos x^2 . (\frac{2x^2}{2x^2}) }{3x^2 \tan x . (\frac{x}{x}) + \frac{x^3}{1 - \sin^2 x . (\frac{x^2}{x^2})}}\right)[/MATH]= [MATH]\lim_{x\to 0} \left(\frac{\frac{2x^3}{\sqrt{1+ x^4}} + 6x^3 \cos x^2 }{3x^3 + \frac{x^3}{1 - x^2}}\right)[/MATH], Algebra:
=[MATH]\lim_{x\to 0} \left(\frac{\frac{2x^3\sqrt{1+x^4}}{1+ x^4} + 6x^3 \cos x^2 }{\frac{3x^3(1-x^2) + x^3}{1 - x^2}}\right)[/MATH]= [MATH]\lim_{x\to 0} \left(\frac{\frac{2x^3\sqrt{1+x^4}+6x^3\cos x^2(1+x^4)}{1+x^4}}{\frac{3x^3(1-x^2) + x^3}{1 - x^2}}\right)[/MATH]= [MATH]\lim_{x\to 0} \left(\frac{\frac{x^3\bigl(2\sqrt{1+x^4}+6\cos x^2(1+x^4)\bigr)}{1+x^4}}{\frac{x^3\bigl(3(1-x^2) + 1 \bigr)}{1 - x^2}}\right)[/MATH]
[MATH]\rightarrow[/MATH] [MATH]x^3[/MATH] cancels, we're left with:
[MATH]\lim_{x\to 0} \left(\frac{\frac{(2\sqrt{1+x^4}+6\cos x^2(1+x^4))}{1+x^4}}{\frac{3(1-x^2) + 1}{1 - x^2}}\right)[/MATH]
[MATH]\rightarrow[/MATH]Now, we plug in [MATH]x = 0[/MATH], result:
[MATH]\left(\frac{\frac{(2\sqrt{1+0^4}+6\cos 0^2(1+0^4))}{1+0^4}}{\frac{3(1-0^2) + 1}{1 - 0^2}}\right)[/MATH]= [MATH]\frac{2+6}{3+1}[/MATH]= [MATH]\frac{8}{4}[/MATH]= [MATH]2[/MATH]
This is how i first got the answer. messy work, i know.
Now that i think about it, i could've changed some unnecessary steps here and there, and it would've looked cleaner.
But still, i don't know if this answer is correct or not, so that's why i'm asking here. If somebody find any mistakes, please let me know.
PS - i'm in my last year of high school, and English is not my native language, so bear with me if i don't understand some mathematical terms.
Evaluate this limit: [MATH]\lim_{x\to 0} \left(\frac{\sqrt{1+\sin^2 x^2} - \cos^3 x^2}{x^3 \tan x}\right)[/MATH]
So, i decided to try and answer the problem. After a few hours, i've found a solution for it, but i feel a bit unsure. Here's my solution:
[MATH]\rightarrow[/MATH] Using L'Hopital's rule,
=[MATH]\lim_{x\to 0} \left(\frac{\frac{2x\sin x^2\cos x^2}{\sqrt{1+\sin^2 x^2}} + 6x\cos^2 x^2 \sin x^2}{3x^2 \tan x + x^3\sec^2 x}\right)[/MATH][MATH]\rightarrow[/MATH]Since [MATH]\sin 2x = 2\sin x \cos x[/MATH], Then [MATH]\sin 2x^2 = 2\sin x^2 \cos x^2[/MATH]
=[MATH]\lim_{x\to 0} \left(\frac{\frac{x\sin 2x^2}{\sqrt{1+\sin^2 x^2}} + 3x\sin 2x^2 \cos x^2 }{3x^2 \tan x + \frac{x^3}{\cos^2 x}}\right)[/MATH][MATH]\rightarrow[/MATH][MATH]\cos^2 x = 1 - \sin^2 x[/MATH]
=[MATH]\lim_{x\to 0} \left(\frac{\frac{x\sin 2x^2}{\sqrt{1+\sin^2 x^2}} + 3x\sin 2x^2 \cos x^2 }{3x^2 \tan x + \frac{x^3}{1 - \sin^2 x}}\right)[/MATH][MATH]\rightarrow[/MATH]Trigonometric limit properties,
=[MATH]\lim_{x\to 0} \left(\frac{\frac{x\sin 2x^2}{\sqrt{(1+\sin^2 x^2) . (\frac{x^4}{x^4})}}. (\frac{2x^2}{2x^2}) + 3x\sin 2x^2 \cos x^2 . (\frac{2x^2}{2x^2}) }{3x^2 \tan x . (\frac{x}{x}) + \frac{x^3}{1 - \sin^2 x . (\frac{x^2}{x^2})}}\right)[/MATH]= [MATH]\lim_{x\to 0} \left(\frac{\frac{2x^3}{\sqrt{1+ x^4}} + 6x^3 \cos x^2 }{3x^3 + \frac{x^3}{1 - x^2}}\right)[/MATH], Algebra:
=[MATH]\lim_{x\to 0} \left(\frac{\frac{2x^3\sqrt{1+x^4}}{1+ x^4} + 6x^3 \cos x^2 }{\frac{3x^3(1-x^2) + x^3}{1 - x^2}}\right)[/MATH]= [MATH]\lim_{x\to 0} \left(\frac{\frac{2x^3\sqrt{1+x^4}+6x^3\cos x^2(1+x^4)}{1+x^4}}{\frac{3x^3(1-x^2) + x^3}{1 - x^2}}\right)[/MATH]= [MATH]\lim_{x\to 0} \left(\frac{\frac{x^3\bigl(2\sqrt{1+x^4}+6\cos x^2(1+x^4)\bigr)}{1+x^4}}{\frac{x^3\bigl(3(1-x^2) + 1 \bigr)}{1 - x^2}}\right)[/MATH]
[MATH]\rightarrow[/MATH] [MATH]x^3[/MATH] cancels, we're left with:
[MATH]\lim_{x\to 0} \left(\frac{\frac{(2\sqrt{1+x^4}+6\cos x^2(1+x^4))}{1+x^4}}{\frac{3(1-x^2) + 1}{1 - x^2}}\right)[/MATH]
[MATH]\rightarrow[/MATH]Now, we plug in [MATH]x = 0[/MATH], result:
[MATH]\left(\frac{\frac{(2\sqrt{1+0^4}+6\cos 0^2(1+0^4))}{1+0^4}}{\frac{3(1-0^2) + 1}{1 - 0^2}}\right)[/MATH]= [MATH]\frac{2+6}{3+1}[/MATH]= [MATH]\frac{8}{4}[/MATH]= [MATH]2[/MATH]
This is how i first got the answer. messy work, i know.
Now that i think about it, i could've changed some unnecessary steps here and there, and it would've looked cleaner.
But still, i don't know if this answer is correct or not, so that's why i'm asking here. If somebody find any mistakes, please let me know.
PS - i'm in my last year of high school, and English is not my native language, so bear with me if i don't understand some mathematical terms.
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