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Find  \mathrm{\lim _{x \rightarrow 0} \frac{1-\cos (a x)}{1-\sqrt{1+x^2}}}

Option: 1

\mathrm{-a^{2}}


Option: 2

\mathrm{a^{2}}


Option: 3

0


Option: 4

None


Answers (1)

best_answer

\mathrm{ \lim _{x \rightarrow 0} \frac{1-\cos (a x)}{1-\sqrt{1+x^2}}=\lim _{x \rightarrow 0} \frac{2 \cdot \frac{1-\cos (a x)}{2}}{1-\sqrt{1+x^2}}=\lim _{x \rightarrow 0} \frac{2 \sin ^2\left(\frac{a x}{2}\right)}{1-\sqrt{1+x^2}} }

Then we proceed to multiply the numerator and denominator by \mathrm{1-\sqrt{1+x^2} :}

\mathrm{ \lim _{x \rightarrow 0} \frac{2 \sin ^2\left(\frac{a x}{2}\right)}{1-\sqrt{1+x^2}} \cdot \frac{1-\sqrt{1+x^2}}{1-\sqrt{1+x^2}}=\lim _{x \rightarrow 0}-\frac{\left[2 \sin ^2\left(\frac{a x}{2}\right)\right]\left[1-\sqrt{1+x^2}\right]}{x^2} }
Then, to be able to use the fact that  \mathrm{\lim _{x \rightarrow 0} \frac{\sin x}{x}=1}, we multiply the numerator and denominator by \mathrm{\frac{a^2}{4} : }

\mathrm{ \lim _{x \rightarrow 0}-\frac{\left[2 \sin ^2\left(\frac{a x}{2}\right)\right]\left[1-\sqrt{1+x^2}\right]}{x^2} \cdot \frac{\frac{a^2}{4}}{\frac{a^2}{4}}=\lim _{x \rightarrow 0} \frac{\left[\sin ^2\left(\frac{a x}{2}\right)\right]\left[-\frac{1}{2} a^2\right]\left[1-\sqrt{1+x^2}\right]}{\frac{a^2 x^2}{4}} }

\mathrm{\begin{aligned} & =\lim _{x \rightarrow 0} \frac{\sin ^2\left(\frac{a x}{2}\right)}{\frac{a^2 x^2}{4}} \cdot \lim _{x \rightarrow 0}\left(-\frac{1}{2} a^2\right)\left(1-\sqrt{1+x^2}\right) \\ & =\left(\lim _{x \rightarrow 0} \frac{\sin \left(\frac{a x}{2}\right)}{\frac{a x}{2}}\right)^2 \cdot \lim _{x \rightarrow 0}\left(-\frac{1}{2} a^2\right)\left(1-\sqrt{1+x^2}\right) \\ & =\lim _{x \rightarrow 0}\left(-\frac{1}{2} a^2\right)\left(1-\sqrt{1+x^2}\right)=-a^2 \end{aligned}}

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