Introduction to limits

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$If \lim_{x \to \infty} x^{α} (\sqrt{x^{2} + \sqrt{x^{4} + 1}} - \sqrt{2} x) exists and has value non-zero finite real number$ $L, then the value of 100 - \frac{α}{L^{2}} is$

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$\begin{aligned} we have lim_{x \to \infty} x^{α} \frac{(x^{2} + \sqrt{x^{4} + 1}) - 2 x^{2}}{{(x^{2} + \sqrt{x^{4} + 1})}^{1 / 2} + \sqrt{2} x} \\ = lim_{x \to \infty} x^{α} \frac{(\sqrt{x^{4} + 1} - x^{2})}{({(x^{2} + \sqrt{x^{4} + 1})}^{1 / 2} + \sqrt{2} x)} \\ = lim_{x \to \infty} x^{α + 1} \frac{(\sqrt{1 + \frac{1}{x^{4}}} - 1)}{{(\sqrt{1 + \frac{1}{x^{4}}} + 1)}^{1 / 2} + \sqrt{2}} \\ = \frac{1}{2 \sqrt{2}} lim_{x \to \infty} x^{α + 1} (1 + \frac{1}{2} \cdot \frac{1}{x^{4}} + \dots - 1) \end{aligned}$
So, for above limit to exist and has value non-zero, we must have
$\begin{array}{l} α + 1 = 4 \Rightarrow α = 3 and L = \frac{1}{2 \sqrt{2}} \times \frac{1}{2} = \frac{1}{4 \sqrt{2}} \\ \Rightarrow L^{2} = \frac{1}{32} \end{array}$

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Introduction to limits

Remember concepts with our Masterclasses.

If limx→∞xαx2+x4+1-2x exists and has value non-zero finite real number L, then the value of 100-αL2 is

we have limx→∞ xαx2+x4+1−2x2x2+x4+11/2+2x=limx→∞ xαx4+1−x2x2+x4+11/2+2x=limx→∞ xα+11+1x4−11+1x4+11/2+2=122limx→∞ xα+11+12⋅1x4+…−1So, for above limit to exist and has value non-zero, we must haveα+1=4⇒α=3 and L=122×12=142⇒L2=132

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$If \lim_{x \to \infty} x^{α} (\sqrt{x^{2} + \sqrt{x^{4} + 1}} - \sqrt{2} x) exists and has value non-zero finite real number$ $L, then the value of 100 - \frac{α}{L^{2}} is$