![Q.55 The right limit limx→3⁺ [log((x-3)) + csc³((x-3)) + (x-3)²] is _____ .](https://www.letstalkacademy.com/wp-content/uploads/2025/12/slide_55-10.jpg)
Q.55
The right limit limx→3⁺ [log((x-3)) + csc³((x-3)) + (x-3)²] is _____ .
Introduction
This article explains step by step how to evaluate the right limit of the function (x−3)²(log(x−3)+csc((x−3)²)) as x approaches 3 from the right.
Understanding this limit combines ideas from logarithmic limits and trigonometric asymptotics, which are important for calculus and competitive exams.
Step 1: Substitution
Start with the limit
\(\lim_{x \to 3^+}(x-3)^2(\log(x-3)+\csc((x-3)^2))\).
Let t = x − 3, so as x → 3⁺, t → 0⁺ and the limit becomes \(\lim_{t \to 0^+} t^2(\log t + \csc(t^2))\).
Step 2: Limit of t² log t
For \(\lim_{t \to 0^+} t^2 \log t\), note that log t → −∞ while t² → 0⁺, giving the indeterminate form 0·(−∞).
Using the standard result \(\lim_{t \to 0^+} t^\alpha \log t = 0\) for any α>0, this term tends to 0, so it does not affect the final value.
Step 3: Limit of t² csc(t²)
Rewrite t²csc(t²) as t²/ sin(t²).
Using the fundamental limit \(\lim_{u \to 0} \sin u / u = 1\), it follows that \(\lim_{t \to 0} t^2/\sin(t^2)=1\), so \(\lim_{t \to 0^+} t^2 \csc(t^2)=1\).
Step 4: Combine the results
The original limit equals \(\lim_{t \to 0^+} t^2 \log t + \lim_{t \to 0^+} t^2 \csc(t^2)\) because each limit exists.
Hence the value is 0 + 1 = 1, so the right–hand limit of (x−3)²(log(x−3)+csc((x−3)²)) as x → 3⁺ is 1.
