\( \begin{array}{lll}\text { (g) } \lim _{x \rightarrow \infty} \frac{\ln x}{\sqrt{x}} & \text { (h) } \lim _{u \rightarrow \infty} \frac{u}{\sqrt{x^{2}+1}} & \text { (i) } \lim _{x \rightarrow \infty} \sqrt{x}(\sqrt{x+3}-\sqrt{x-2}) \\ \text { (j) } \lim _{x \rightarrow \infty} \frac{3 x+|1-3 x|}{1-5 x} & \text { (k) } \lim _{x \rightarrow \infty}(\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}) & \text { (1) } \lim _{x \rightarrow-\infty} \frac{\sqrt{x^{2}+4 x}}{4 x+1} \\ \text { (m) } \lim _{x \rightarrow \infty} \frac{\sqrt{3 x^{8}+7 x^{4}}+10}{x^{4}-2 x^{2}+1} & \text { (n) } \lim _{x \rightarrow \infty}\left[\sqrt{x^{2}+5 x}-\sqrt{x^{2}-x}\right] & \text { (o) } \lim _{x \rightarrow \infty} \frac{3 x}{\sqrt{4 x^{2}+x}-2 x}\end{array} \)

Evaluate the limit by following steps: - step0: Evaluate the limit: \(\lim _{x\rightarrow +\infty}\left(\frac{3x+\left|1-3x\right|}{1-5x}\right)\) - step1: Rewrite the expression: \(\lim _{x\rightarrow +\infty}\left(\frac{3x-1+3x}{1-5x}\right)\) - step2: Simplify: \(\lim _{x\rightarrow +\infty}\left(\frac{6x-1}{1-5x}\right)\) - step3: Rewrite the expression: \(\lim _{x\rightarrow +\infty}\left(\frac{x\left(6-\frac{1}{x}\right)}{x\left(\frac{1}{x}-5\right)}\right)\) - step4: Reduce the fraction: \(\lim _{x\rightarrow +\infty}\left(\frac{6-\frac{1}{x}}{\frac{1}{x}-5}\right)\) - step5: Rewrite the expression: \(\frac{\lim _{x\rightarrow +\infty}\left(6-\frac{1}{x}\right)}{\lim _{x\rightarrow +\infty}\left(\frac{1}{x}-5\right)}\) - step6: Calculate: \(\frac{6}{\lim _{x\rightarrow +\infty}\left(\frac{1}{x}-5\right)}\) - step7: Calculate: \(\frac{6}{-5}\) - step8: Calculate: \(-\frac{6}{5}\) Calculate the limit \( \lim_{x \rightarrow \infty} \frac{3 x}{\sqrt{4 x^{2}+x}-2 x} \). Evaluate the limit by following steps: - step0: Evaluate using transformations: \(\lim _{x\rightarrow +\infty}\left(\frac{3x}{\sqrt{4x^{2}+x}-2x}\right)\) - step1: Rewrite the expression: \(\frac{\lim _{x\rightarrow +\infty}\left(3x\right)}{\lim _{x\rightarrow +\infty}\left(\sqrt{4x^{2}+x}-2x\right)}\) - step2: Calculate: \(\frac{+\infty}{\lim _{x\rightarrow +\infty}\left(\sqrt{4x^{2}+x}-2x\right)}\) - step3: Calculate: \(\frac{+\infty}{\frac{1}{4}}\) - step4: Calculate: \(+\infty\) Calculate the limit \( \lim_{x \rightarrow \infty} \frac{\ln x}{\sqrt{x}} \). Evaluate the limit by following steps: - step0: Evaluate using L'Hopital's rule: \(\lim _{x\rightarrow +\infty}\left(\frac{\ln{\left(x\right)}}{\sqrt{x}}\right)\) - step1: Use the L'Hopital's rule: \(\lim _{x\rightarrow +\infty}\left(\frac{\frac{d}{dx}\left(\ln{\left(x\right)}\right)}{\frac{d}{dx}\left(\sqrt{x}\right)}\right)\) - step2: Find the derivative: \(\lim _{x\rightarrow +\infty}\left(\frac{\frac{1}{x}}{\frac{1}{2\sqrt{x}}}\right)\) - step3: Simplify the expression: \(\lim _{x\rightarrow +\infty}\left(\frac{2}{\sqrt{x}}\right)\) - step4: Rewrite the expression: \(\frac{2}{\lim _{x\rightarrow +\infty}\left(\sqrt{x}\right)}\) - step5: Calculate: \(\frac{2}{+\infty}\) - step6: Divide the terms: \(0\) Calculate the limit \( \lim_{x \rightarrow \infty} \sqrt{x}(\sqrt{x+3}-\sqrt{x-2}) \). Evaluate the limit by following steps: - step0: Evaluate using transformations: \(\lim _{x\rightarrow +\infty}\left(\sqrt{x}\times \left(\sqrt{x+3}-\sqrt{x-2}\right)\right)\) - step1: Calculate the product: \(\lim _{x\rightarrow +\infty}\left(\sqrt{x^{2}+3x}-\sqrt{x^{2}-2x}\right)\) - step2: Multiply by the Conjugate: \(\lim _{x\rightarrow +\infty}\left(\frac{5x}{\sqrt{x^{2}+3x}+\sqrt{x^{2}-2x}}\right)\) - step3: Rearrange the terms: \(\lim _{x\rightarrow +\infty}\left(\frac{5x}{\left(\sqrt{1+\frac{3}{x}}+\sqrt{1-\frac{2}{x}}\right)x}\right)\) - step4: Reduce the fraction: \(\lim _{x\rightarrow +\infty}\left(\frac{5}{\sqrt{1+\frac{3}{x}}+\sqrt{1-\frac{2}{x}}}\right)\) - step5: Rewrite the expression: \(\frac{\lim _{x\rightarrow +\infty}\left(5\right)}{\lim _{x\rightarrow +\infty}\left(\sqrt{1+\frac{3}{x}}+\sqrt{1-\frac{2}{x}}\right)}\) - step6: Evaluate: \(\frac{5}{\lim _{x\rightarrow +\infty}\left(\sqrt{1+\frac{3}{x}}+\sqrt{1-\frac{2}{x}}\right)}\) - step7: Evaluate: \(\frac{5}{2}\) Calculate the limit \( \lim_{x \rightarrow -\infty} \frac{\sqrt{x^{2}+4 x}}{4 x+1} \). Evaluate the limit by following steps: - step0: Evaluate using transformations: \(\lim _{x\rightarrow -\infty}\left(\frac{\sqrt{x^{2}+4x}}{4x+1}\right)\) - step1: Rewrite the expression: \(\lim _{x\rightarrow -\infty}\left(\frac{-\sqrt{1+\frac{4}{x}}\times x}{\left(4+\frac{1}{x}\right)x}\right)\) - step2: Reduce the fraction: \(\lim _{x\rightarrow -\infty}\left(\frac{-\sqrt{1+\frac{4}{x}}}{4+\frac{1}{x}}\right)\) - step3: Rewrite the expression: \(\frac{\lim _{x\rightarrow -\infty}\left(-\sqrt{1+\frac{4}{x}}\right)}{\lim _{x\rightarrow -\infty}\left(4+\frac{1}{x}\right)}\) - step4: Calculate: \(\frac{-1}{\lim _{x\rightarrow -\infty}\left(4+\frac{1}{x}\right)}\) - step5: Calculate: \(\frac{-1}{4}\) - step6: Rewrite the fraction: \(-\frac{1}{4}\) Calculate the limit \( \lim_{x \rightarrow \infty} \frac{\sqrt{3 x^{8}+7 x^{4}}+10}{x^{4}-2 x^{2}+1} \). Evaluate the limit by following steps: - step0: Evaluate using transformations: \(\lim _{x\rightarrow +\infty}\left(\frac{\sqrt{3x^{8}+7x^{4}}+10}{x^{4}-2x^{2}+1}\right)\) - step1: Simplify the root: \(\lim _{x\rightarrow +\infty}\left(\frac{x^{2}\sqrt{3x^{4}+7}+10}{x^{4}-2x^{2}+1}\right)\) - step2: Rewrite the expression: \(\lim _{x\rightarrow +\infty}\left(\frac{\left(\sqrt{3+\frac{7}{x^{4}}}+\frac{10}{x^{4}}\right)x^{4}}{\left(1-\frac{2}{x^{2}}+\frac{1}{x^{4}}\right)x^{4}}\right)\) - step3: Reduce the fraction: \(\lim _{x\rightarrow +\infty}\left(\frac{\sqrt{3+\frac{7}{x^{4}}}+\frac{10}{x^{4}}}{1-\frac{2}{x^{2}}+\frac{1}{x^{4}}}\right)\) - step4: Rewrite the expression: \(\frac{\lim _{x\rightarrow +\infty}\left(\sqrt{3+\frac{7}{x^{4}}}+\frac{10}{x^{4}}\right)}{\lim _{x\rightarrow +\infty}\left(1-\frac{2}{x^{2}}+\frac{1}{x^{4}}\right)}\) - step5: Calculate: \(\frac{\sqrt{3}}{\lim _{x\rightarrow +\infty}\left(1-\frac{2}{x^{2}}+\frac{1}{x^{4}}\right)}\) - step6: Calculate: \(\frac{\sqrt{3}}{1}\) - step7: Divide the terms: \(\sqrt{3}\) Calculate the limit \( \lim_{x \rightarrow \infty}\left[\sqrt{x^{2}+5 x}-\sqrt{x^{2}-x}\right] \). Evaluate the limit by following steps: - step0: Evaluate using transformations: \(\lim _{x\rightarrow +\infty}\left(\sqrt{x^{2}+5x}-\sqrt{x^{2}-x}\right)\) - step1: Multiply by the Conjugate: \(\lim _{x\rightarrow +\infty}\left(\frac{6x}{\sqrt{x^{2}+5x}+\sqrt{x^{2}-x}}\right)\) - step2: Rearrange the terms: \(\lim _{x\rightarrow +\infty}\left(\frac{6x}{\left(\sqrt{1+\frac{5}{x}}+\sqrt{1-\frac{1}{x}}\right)x}\right)\) - step3: Reduce the fraction: \(\lim _{x\rightarrow +\infty}\left(\frac{6}{\sqrt{1+\frac{5}{x}}+\sqrt{1-\frac{1}{x}}}\right)\) - step4: Rewrite the expression: \(\frac{\lim _{x\rightarrow +\infty}\left(6\right)}{\lim _{x\rightarrow +\infty}\left(\sqrt{1+\frac{5}{x}}+\sqrt{1-\frac{1}{x}}\right)}\) - step5: Evaluate: \(\frac{6}{\lim _{x\rightarrow +\infty}\left(\sqrt{1+\frac{5}{x}}+\sqrt{1-\frac{1}{x}}\right)}\) - step6: Evaluate: \(\frac{6}{2}\) - step7: Reduce the numbers: \(\frac{3}{1}\) - step8: Calculate: \(3\) Calculate the limit \( \lim_{x \rightarrow \infty}(\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}) \). Evaluate the limit by following steps: - step0: Evaluate using transformations: \(\lim _{x\rightarrow +\infty}\left(\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}\right)\) - step1: Multiply by the Conjugate: \(\lim _{x\rightarrow +\infty}\left(\frac{2\sqrt{x}}{\sqrt{x+\sqrt{x}}+\sqrt{x-\sqrt{x}}}\right)\) - step2: Rearrange the terms: \(\lim _{x\rightarrow +\infty}\left(\frac{2x^{\frac{1}{2}}}{\left(\sqrt{1+\sqrt{x^{-1}}}+\sqrt{1-\sqrt{x^{-1}}}\right)x^{\frac{1}{2}}}\right)\) - step3: Reduce the fraction: \(\lim _{x\rightarrow +\infty}\left(\frac{2}{\sqrt{1+\sqrt{x^{-1}}}+\sqrt{1-\sqrt{x^{-1}}}}\right)\) - step4: Rewrite the expression: \(\frac{\lim _{x\rightarrow +\infty}\left(2\right)}{\lim _{x\rightarrow +\infty}\left(\sqrt{1+\sqrt{x^{-1}}}+\sqrt{1-\sqrt{x^{-1}}}\right)}\) - step5: Evaluate: \(\frac{2}{\lim _{x\rightarrow +\infty}\left(\sqrt{1+\sqrt{x^{-1}}}+\sqrt{1-\sqrt{x^{-1}}}\right)}\) - step6: Evaluate: \(\frac{2}{2}\) - step7: Reduce the numbers: \(\frac{1}{1}\) - step8: Calculate: \(1\) Let's evaluate each limit step by step: ### (g) \( \lim_{x \rightarrow \infty} \frac{\ln x}{\sqrt{x}} \) As \( x \) approaches infinity, \( \ln x \) grows much slower than \( \sqrt{x} \). Therefore, the limit is: \[ \lim_{x \rightarrow \infty} \frac{\ln x}{\sqrt{x}} = 0 \] ### (h) \( \lim_{u \rightarrow \infty} \frac{u}{\sqrt{x^{2}+1}} \) This limit is not well-defined as it contains \( x \) instead of \( u \). Assuming \( x \) is a constant, the limit cannot be evaluated without further information. ### (i) \( \lim_{x \rightarrow \infty} \sqrt{x}(\sqrt{x+3}-\sqrt{x-2}) \) Using the identity \( a - b = \frac{(a^2 - b^2)}{a + b} \): \[ \sqrt{x+3} - \sqrt{x-2} = \frac{(x+3) - (x-2)}{\sqrt{x+3} + \sqrt{x-2}} = \frac{5}{\sqrt{x+3} + \sqrt{x-2}} \] Thus, \[ \lim_{x \rightarrow \infty} \sqrt{x}(\sqrt{x+3}-\sqrt{x-2}) = \lim_{x \rightarrow \infty} \sqrt{x} \cdot \frac{5}{\sqrt{x+3} + \sqrt{x-2}} = \frac{5}{2} \] ### (j) \( \lim_{x \rightarrow \infty} \frac{3 x + |1 - 3 x|}{1 - 5 x} \) For large \( x \), \( |1 - 3x| = 3x - 1 \): \[ \lim_{x \rightarrow \infty} \frac{3x + (3x - 1)}{1 - 5x} = \lim_{x \rightarrow \infty} \frac{6x - 1}{1 - 5x} = \lim_{x \rightarrow \infty} \frac{6 - \frac{1}{x}}{-5} = -\frac{6}{5} \] ### (k) \( \lim_{x \rightarrow \infty}(\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}) \) Using the same identity: \[ \sqrt{x+\sqrt{x}} - \sqrt{x-\sqrt{x}} = \frac{(x+\sqrt{x}) - (x-\sqrt{x})}{\sqrt{x+\sqrt{x}} + \sqrt{x-\sqrt{x}}} = \frac{2\sqrt{x}}{\sqrt{x+\sqrt{x}} + \sqrt{x-\sqrt{x}}} \] As \( x \) approaches infinity, this limit evaluates to: \[ \lim_{x \rightarrow \infty} \frac{2\sqrt{x}}{2\sqrt{x}} = 1 \] ### (1) \( \lim_{x \rightarrow -\infty} \frac{\sqrt{x^{2}+4 x}}{4 x+1} \) As \( x \) approaches negative infinity: \[ \lim_{x \rightarrow -\infty} \frac{\sqrt{x^2(1 + \frac{4}{x})}}{4x(1 + \frac{1}{4x})} = \lim_{x \rightarrow -\infty} \frac{|x|\sqrt{1 + \frac{4}{x}}}{4x(1 + \frac{1}{4x})} = \lim_{x \rightarrow -\infty} \frac{-\sqrt{1 + \frac{4}{x}}}{4(1 + \frac{1}{4x})} = -\frac{1}{4} \] ### (m) \( \lim_{x \rightarrow \infty} \frac{\sqrt{3 x^{8}+7 x^{4}}+10}{x^{4}-2 x^{2}+1} \) The dominant term in the numerator is \( \sqrt{3}x^4 \) and in the denominator is \( x^4 \): \[ \lim_{x \rightarrow \infty} \frac{\sqrt{3}x^4 + 10}{x^4 - 2x^2 + 1} = \sqrt{3} \] ### (n) \( \lim_{x \rightarrow \infty}\left[\sqrt{x^{2}+5 x}-\sqrt{x^{2}-x}\right] \) Using the identity: \[ \sqrt{x^2 + 5x} - \sqrt{x^2 - x} = \frac{(x^2 + 5x) - (x^2 - x)}{\sqrt{x^2 + 5x} + \sqrt{x^2 - x}} = \frac{6x}{\sqrt{x^2 + 5x} + \sqrt{x^2 - x}} \] As \( x \) approaches infinity, this limit evaluates to: \[ \lim_{x \rightarrow \infty} \frac{6x}{2x} = 3 \] ### (o) \( \lim_{x \rightarrow \infty} \frac{3 x}{\sqrt{4 x^{2}+x}-2 x} \) Rationalizing the denominator: \[ \lim_{x \rightarrow \infty} \frac{3x(\sqrt{4x^2 + x} + 2x)}{(4x^2 + x) - 4x^2} = \lim_{x \rightarrow \infty} \frac{3x(\sqrt{4x^2 + x} + 2x)}{x} = 3 \] ### Summary of Results: - (g) \( 0 \) - (h) Undefined - (i) \( \frac{5}{2} \) - (j) \( -\frac{6}{5} \) - (k) \( 1 \) - (1) \( -\frac{1}{4} \) - (m) \( \sqrt{3} \) - (n) \( 3 \) - (o) \( 3 \)