79915 If \(f(x)=\left\{\begin{array}{cc}\log _{(1-3 x)}(1+3 x), x \neq 0 \\ k, x=0\end{array}\right.\) is continuous at \(x=0\), then value of \(k\) equals

79916 If \(f(x)\) is continuous over \([-\pi, \pi]\) where \(f(x)\) is defined as
\(f(x)= \begin{cases}-2 \sin x, -\pi \leq x \leq \frac{-\pi}{2} \\ \alpha \sin x+\beta, -\frac{\pi}{2}\lt x\lt \frac{\pi}{2} \\ \cos x, \frac{\pi}{2} \leq x \leq \pi\end{cases}\)
then \(\alpha\) and \(\beta\) equals

79917 \(\lim _{x \rightarrow 0}\left(\frac{1+5 x^{2}}{1+3 x^{2}}\right)^{\frac{1}{x^{2}}}=\)

79918 \(\lim _{x \rightarrow 0} \frac{(1-\cos 2 x)(3+\cos x)}{x \tan 4 x}\) is equal to

79919 \(\underset{x \rightarrow 0}{\operatorname{Lt}} \frac{x \cdot a^{x}-x}{1-\cos x}\) is equal to

79915 If \(f(x)=\left\{\begin{array}{cc}\log _{(1-3 x)}(1+3 x), x \neq 0 \\ k, x=0\end{array}\right.\) is continuous at \(x=0\), then value of \(k\) equals

79916 If \(f(x)\) is continuous over \([-\pi, \pi]\) where \(f(x)\) is defined as
\(f(x)= \begin{cases}-2 \sin x, -\pi \leq x \leq \frac{-\pi}{2} \\ \alpha \sin x+\beta, -\frac{\pi}{2}\lt x\lt \frac{\pi}{2} \\ \cos x, \frac{\pi}{2} \leq x \leq \pi\end{cases}\)
then \(\alpha\) and \(\beta\) equals

79917 \(\lim _{x \rightarrow 0}\left(\frac{1+5 x^{2}}{1+3 x^{2}}\right)^{\frac{1}{x^{2}}}=\)

79918 \(\lim _{x \rightarrow 0} \frac{(1-\cos 2 x)(3+\cos x)}{x \tan 4 x}\) is equal to

79919 \(\underset{x \rightarrow 0}{\operatorname{Lt}} \frac{x \cdot a^{x}-x}{1-\cos x}\) is equal to

79915 If \(f(x)=\left\{\begin{array}{cc}\log _{(1-3 x)}(1+3 x), x \neq 0 \\ k, x=0\end{array}\right.\) is continuous at \(x=0\), then value of \(k\) equals

79916 If \(f(x)\) is continuous over \([-\pi, \pi]\) where \(f(x)\) is defined as
\(f(x)= \begin{cases}-2 \sin x, -\pi \leq x \leq \frac{-\pi}{2} \\ \alpha \sin x+\beta, -\frac{\pi}{2}\lt x\lt \frac{\pi}{2} \\ \cos x, \frac{\pi}{2} \leq x \leq \pi\end{cases}\)
then \(\alpha\) and \(\beta\) equals

79917 \(\lim _{x \rightarrow 0}\left(\frac{1+5 x^{2}}{1+3 x^{2}}\right)^{\frac{1}{x^{2}}}=\)

79918 \(\lim _{x \rightarrow 0} \frac{(1-\cos 2 x)(3+\cos x)}{x \tan 4 x}\) is equal to

79919 \(\underset{x \rightarrow 0}{\operatorname{Lt}} \frac{x \cdot a^{x}-x}{1-\cos x}\) is equal to

79915 If \(f(x)=\left\{\begin{array}{cc}\log _{(1-3 x)}(1+3 x), x \neq 0 \\ k, x=0\end{array}\right.\) is continuous at \(x=0\), then value of \(k\) equals

79916 If \(f(x)\) is continuous over \([-\pi, \pi]\) where \(f(x)\) is defined as
\(f(x)= \begin{cases}-2 \sin x, -\pi \leq x \leq \frac{-\pi}{2} \\ \alpha \sin x+\beta, -\frac{\pi}{2}\lt x\lt \frac{\pi}{2} \\ \cos x, \frac{\pi}{2} \leq x \leq \pi\end{cases}\)
then \(\alpha\) and \(\beta\) equals

79917 \(\lim _{x \rightarrow 0}\left(\frac{1+5 x^{2}}{1+3 x^{2}}\right)^{\frac{1}{x^{2}}}=\)

79918 \(\lim _{x \rightarrow 0} \frac{(1-\cos 2 x)(3+\cos x)}{x \tan 4 x}\) is equal to

79919 \(\underset{x \rightarrow 0}{\operatorname{Lt}} \frac{x \cdot a^{x}-x}{1-\cos x}\) is equal to

79915 If \(f(x)=\left\{\begin{array}{cc}\log _{(1-3 x)}(1+3 x), x \neq 0 \\ k, x=0\end{array}\right.\) is continuous at \(x=0\), then value of \(k\) equals

79916 If \(f(x)\) is continuous over \([-\pi, \pi]\) where \(f(x)\) is defined as
\(f(x)= \begin{cases}-2 \sin x, -\pi \leq x \leq \frac{-\pi}{2} \\ \alpha \sin x+\beta, -\frac{\pi}{2}\lt x\lt \frac{\pi}{2} \\ \cos x, \frac{\pi}{2} \leq x \leq \pi\end{cases}\)
then \(\alpha\) and \(\beta\) equals

79917 \(\lim _{x \rightarrow 0}\left(\frac{1+5 x^{2}}{1+3 x^{2}}\right)^{\frac{1}{x^{2}}}=\)

79918 \(\lim _{x \rightarrow 0} \frac{(1-\cos 2 x)(3+\cos x)}{x \tan 4 x}\) is equal to

79919 \(\underset{x \rightarrow 0}{\operatorname{Lt}} \frac{x \cdot a^{x}-x}{1-\cos x}\) is equal to