Lim as x-->infinity of (x + sqrt(x^2 +2X))?
[ 1 ] Ans. : This limit does NOT exist ......................................... Mathematical Explanation ......................................... Rationalizing, the given function becomes [ x^2 - ( x^2 + 2x ) ] / sqrt [ x - sqrt ( x^2 + 2x ) ] = - 2x / [ x - sqrt ( x^2 + 2x ) ]......divide N and D by x = rt ( x^2 ) = - 2 / [ 1 - sqrt ( 1 + 2/x ) ] ......................................... As x --> inf., the limit of this function becomes L = - 2 / [ 1 - ( 1 + 0 ) ] = - 2 / 0, i.e., -infinity, which does not exist. ......................................... Hence, the req'd limit does not exist. ......................................... [ 2 ] After rationalization, this function becomes = x / [ sqrt( x^2 + x ) + x ].....divide N and D by x = rt ( x^2 ) = 1 / [ sqrt( 1 + 1/x ) + 1 ]. ......................................... As x --> inf., its limit becomes L = 1 / [ sqrt( 1 + 0 ) + 1 ] = 1 / ( 1 + 1 ) = 1 / 2. ................................Ans. ......................................... Happy To Help !
Calculus Help, again! limits! As x -> infinity (sq rt(x^2-9))/(2x-6)?
Analyse the end-run behavior of the function. Note that for large x, the -9 term under the radical is insignificant. That is the x² term dominates. Similarly, in the denominator, the -6 is also insignificant because the 2x term dominates. Thus, for large x, the function behaves like √(x²) / 2x = x / 2x = 1/2 Thus, we have lim x → ∞ √(x² - 9) / 2x - 6 = 1/2
What is the limit: lim x->infinity (2x)/sqrt(x^2+1) ? Can you show me the steps?
lim = 2x/x = 2
Lim x->infinity?
Eval, Lim x->Infin 3x/(sqRt(4x^2+10)) So I take the derivative of the top and bottom right? bottom deriv is .5(4x^2+10)^-.5 * 8x. I think... next step i'm not sure what to do. switch it to the top?
What is the answer of limit x tends to infinity sqrt ((x - sinx) / (x + cos^2x)?
[math]L = \lim \limits_{x\to \infty} \sqrt{\frac{x-\sin x}{x+ \cos^2 x}}.[/math]Let [math]x=\frac{1}{y} \qquad \Rightarrow \qquad[/math] As [math]x\to \infty, \qquad y\to 0.[/math][math]\Rightarrow \qquad L = \lim \limits_{y\to 0} \sqrt{\frac{\frac{1}{y}-\sin \left(\frac{1}{y}\right)}{\frac{1}{y}+ \cos^2 \left(\frac{1}{y}\right)}} = \lim \limits_{y\to 0} \sqrt{\frac{1-y\sin \left(\frac{1}{y}\right)}{1+ y\cos^2 \left(\frac{1}{y}\right)}}.[/math][math]\sin \left(\frac{1}{y}\right)[/math] and [math]\cos^2 \left(\frac{1}{y}\right)[/math] are both bounded functions.[math]\sin \left(\frac{1}{y}\right) \,\,\in \,\, [-1,1] \,\, \forall \,\,y\,\,\in \,\,R[/math] and [math]\cos^2 \left(\frac{1}{y}\right) \,\,\in \,\, [0,1] \,\, \forall \,\,y\,\,\in \,\,R.[/math][math]\Rightarrow \qquad \lim \limits_{y\to 0} y\sin \left(\frac{1}{y}\right)=0[/math] and [math]\lim \limits_{y\to 0} y\cos^2 \left(\frac{1}{y}\right)=0.[/math][math]\Rightarrow \qquad \lim \limits_{y\to 0} \sqrt{\frac{1-y\sin \left(\frac{1}{y}\right)}{1+ y\cos^2 \left(\frac{1}{y}\right)}}=\lim \limits_{y\to 0} \sqrt{\frac{1-0}{1+ 0}} = 1.[/math][math]\Rightarrow \qquad \lim \limits_{x\to \infty} \sqrt{\frac{x-\sin x}{x+ \cos^2 x}}=1.[/math]
What is the limit as x approaches infinity of sqrt(4x^2+x)-2x?
infinity