Chapter 1: Problem 62
Evaluate, \(\operatorname{Lim}_{x \rightarrow 1} \frac{1-x+\ell n x}{1+\cos \pi x}\)
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Chapter 1: Problem 62
Evaluate, \(\operatorname{Lim}_{x \rightarrow 1} \frac{1-x+\ell n x}{1+\cos \pi x}\)
These are the key concepts you need to understand to accurately answer the question.
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The value of \(\lim _{x \rightarrow \infty} \frac{x^{3} \sin (1 / x)-2 x^{2}}{1+3 x^{2}}\) (a) is 0 (b) is \(-1 / 3\) (c) is \(-1\) (d) is \(-2 / 3\)
Roots of equation \(a x^{2}+b x+c=0\) are (a) Real and unequal (b) Imaginary (c) Rational (d) Irrational
If \(\lim _{n \rightarrow \infty} \frac{\sqrt{n^{3}-2 n^{2}+1}+\sqrt[3]{n^{4}+1}}{\sqrt[4]{n^{6}+6 n^{5}+2-\sqrt[5]{n^{7}+3 n^{3}+1}}}=k\), then evaluate \((k)^{2013}\)
1\. \(\operatorname{Lt}_{x \rightarrow 0} \frac{\sin x^{4}-x^{4} \cos x^{4}+x^{20}}{x^{4}\left(e^{2 x^{4}}-1-2 x^{4}\right)}\) is equal to (a) 0 (b) \(1 / 6\) (c) \(1 / 12\) (d) does not exist
\(\begin{aligned}&\text { A: } & \operatorname{Lim}_{x \rightarrow 0} \frac{x-\sin x}{x^{3}}=\operatorname{Lim}_{x \rightarrow 0}\left(\frac{1}{x^{2}}-\frac{\sin x}{x} \cdot \frac{1}{x^{2}}\right) \\\& & =\operatorname{Lim}_{x \rightarrow 0}\left(\frac{1}{x^{2}}-\frac{1}{x^{2}}\right)=0 \\\&\text { R: } & \operatorname{Lim} \frac{\sin x}{x}=1\end{aligned}\)
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