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lim⁡x→∞2x4−74x8+7x5\lim _{x\to \infty }\frac{2x^4-7}{\sqrt{4x^8+7x^5}}

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Q. lim⁡x→∞2x4−74x8+7x5\lim _{x\to \infty }\frac{2x^4-7}{\sqrt{4x^8+7x^5}}
  1. Analyze highest power terms: We start by analyzing the highest power terms in the numerator and the denominator to simplify the expression.\newlinelim⁡x→∞2x4−74x8+7x5 \lim _{x\to \infty }\frac{2x^4-7}{\sqrt{4x^8+7x^5}} \newlineFocus on 2x42x^4 in the numerator and 4x8\sqrt{4x^8} in the denominator.
  2. Simplify square root: Simplify the square root in the denominator:\newline4x8=2x4 \sqrt{4x^8} = 2x^4 \newlineNow, the expression becomes:\newline2x4−72x4 \frac{2x^4-7}{2x^4}
  3. Divide by 22x^44: Divide each term in the numerator by 2x42x^4:\newline2x42x4−72x4=1−72x4 \frac{2x^4}{2x^4} - \frac{7}{2x^4} = 1 - \frac{7}{2x^4}
  4. Evaluate limit: As xx approaches infinity, 72x4\frac{7}{2x^4} approaches 00:\newlinelim⁡x→∞(1−72x4)=1−0=1 \lim _{x\to \infty } \left(1 - \frac{7}{2x^4}\right) = 1 - 0 = 1

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