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\( \begin{array}{lll}\text { (j) } \lim _{x \rightarrow 0} \frac{1}{\sqrt{4-x}-2} & \text { (k) } \lim _{x \rightarrow 0} \frac{1}{x^{2}+3 x} & \text { (1) } \lim _{x \rightarrow 0}(1-2 x)^{x} \\ \text { (m) } \lim _{x \rightarrow 5} \frac{2 w^{2}-50}{(w-5)(w-1)} & \text { (n) } \lim _{x \rightarrow-1} \sqrt{\frac{x^{3}+x^{2}+x+1}{3 x+3}} & \text { (o) } \lim _{x \rightarrow 0^{-}}\left(\frac{1}{x}-\frac{1}{1-2}\right.\end{array} \)

Ask by Daniel Wade. in South Africa
Mar 12,2025

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Here are the results for each limit: 1. **Limit (j)**: Does not exist. 2. **Limit (k)**: Does not exist. 3. **Limit (1)**: 1. 4. **Limit (m)**: 5. 5. **Limit (n)**: \(\frac{\sqrt{6}}{3}\). 6. **Limit (o)**: \(-\infty\).

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To tackle the limit \( \lim _{x \rightarrow 0} \frac{1}{\sqrt{4-x}-2} \), you can rationalize the denominator! Multiply the numerator and denominator by the conjugate \( \sqrt{4-x}+2 \). This gives you a simpler form to evaluate the limit as \( x \) approaches 0, where after some algebraic simplification, things start to clear up! When dealing with limits like \( \lim _{x \rightarrow 0} \frac{1}{x^{2}+3 x} \), it's important to remember that direct substitution may lead to division by zero. Instead, factor out an \( x \) from the denominator, allowing you to see the limit's behavior as \( x \) approaches zero, helping to avoid common pitfalls like overlooking indeterminate forms!

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