Question:
Evaluate the Given limit: \(\lim_{x\rightarrow 3}\) \(\frac{x^4-81}{2x^2-5x-3}\)
Evaluate the Given limit: \(\lim_{x\rightarrow 3}\) \(\frac{x^4-81}{2x^2-5x-3}\)
Updated On: Oct 23, 2023
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Solution and Explanation
At x=2, the value of the given rational function takes the form 0/0
=\(\lim_{x\rightarrow 3}\) \(\frac{(x-3)(x+3)(x^2+9)}{(x-3)(2x+1)}\)
=\(\lim_{x\rightarrow 3}\) \(\frac{(x+3)(x^2+9)}{2x+1}\)
=\(\frac{(3+3)(32+9)}{2(3)+1}\)
=\(\frac{6\times 18}{7}\)
=\(\frac{108}{7}\)
=\(\lim_{x\rightarrow 3}\) \(\frac{(x-3)(x+3)(x^2+9)}{(x-3)(2x+1)}\)
=\(\lim_{x\rightarrow 3}\) \(\frac{(x+3)(x^2+9)}{2x+1}\)
=\(\frac{(3+3)(32+9)}{2(3)+1}\)
=\(\frac{6\times 18}{7}\)
=\(\frac{108}{7}\)
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