$\lim _{x \rightarrow 0} \frac{x+\log (1-x)}{x^{2}}$ equals
$\lim _{x \rightarrow 0} \frac{x+\log (1-x)}{x^{2}}$ = . . . . . .
If $\lim _{x \rightarrow 0} \frac{5\left(\sin x-x+\frac{1}{6} x^{3}\right)}{x^{5}}=\frac{1}{k}$ then $\mathrm{k}$ equals:
यदि $\lim _{x \rightarrow 0} \frac{5\left(\sin x-x+\frac{1}{6} x^{3}\right)}{x^{5}}=\frac{1}{k}$ तो $\mathrm{k}$ का मान क्या होगा ?
$\operatorname{Lim}_{t\rightarrow 0} \frac{\log (1+2 t)-2 \log (1+t)}{t^{2}}=\ldots .$
$\operatorname{Lt}_{x \rightarrow 1}$ $(1-x) \tan \frac{\pi x}{2}=\ldots .$
$\operatorname{Lim}_{x \rightarrow 0} \frac{5^{x}-3^{x}}{3^{x}-2^{x}}=\ldots \ldots$
$\operatorname{Lim}_{x \rightarrow 0} \frac{e^{x^{2}}-\cos x}{x^{2}}$ equals:
$\operatorname{Lim}_{x \rightarrow 0} \frac{e^{x^{2}}-\cos x}{x^{2}}$ = . . . . . . .
$\operatorname{Lim}_{\theta \rightarrow \tan ^{-1} 2} \quad \frac{\left(\tan ^{2} \theta-2 \tan \theta-3\right)}{\left(\tan ^{2} \theta -4 \tan \theta+3\right)}$ equals:
$\operatorname{Lim}_{\theta \rightarrow \tan ^{-1} 2} \quad \frac{\left(\tan ^{2} \theta-2 \tan \theta-3\right)}{\left(\tan ^{2} \theta -4 \tan \theta+3\right)}$ =
$\operatorname{Lim}_{t\rightarrow 0} \frac{|t|}{\sqrt{t^{4}+4 t^{2}+9}}$ equals:
$\operatorname{Lim}_{t\rightarrow 0} \frac{|t|}{\sqrt{t^{4}+4 t^{2}+9}}$ = . . . . . . . . .
$\operatorname{Lim}_{y\rightarrow 0} \frac{\left(1+y^{2}\right)^{\frac{1}{6}}-\left(1-y^{2}\right)^{\frac{1}{6}}}{y^{2}}=\ldots \ldots$
$\operatorname{Lim}_{x \rightarrow 0} \frac{2\left(1-\cos ^{3} x\right)}{x \cdot \sin 2 x}$ is equals to:
$\operatorname{Lim}_{x \rightarrow 0} \frac{2\left(1-\cos ^{3} x\right)}{x \cdot \sin 2 x}$ का मान किसके बराबर होगा ?
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