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Calculus Level 3

$\large \displaystyle \lim_{x \rightarrow 0} \dfrac{e^{x^2} - \cos x}{\sin^2 x } = \, ?$

\dfrac{3}{2}

3

2

None of these choices

\dfrac{5}{4}

1 solution

Rishabh Jain
Mar 9, 2016

$\mathfrak{L}=\lim_{x\to 0}\dfrac{e^{x^2}-1+1-\cos x}{\color{#D61F06}{\sin^2 x}}$ $=\left(\lim_{x\to 0}\dfrac{e^{x^2}-1}{x^2}\right)\left(\lim_{x\to 0}\dfrac{x^2}{\sin^2 x}\right)+\left(\lim_{x\to 0}\dfrac{1-\cos x}{x^2}\right)\left(\lim_{x\to 0}\dfrac{x^2}{\sin^2 x}\right)$ (Using $\lim_{x\to 0}\dfrac{e^x-1}{x}=1, \lim_{x\to 0}\dfrac{x}{\sin x}=1,\lim_{x\to 0}\dfrac{1-\cos x}{x^2}=\dfrac{1}{2}$ ) $\mathfrak{L}=1+\dfrac 12=\boxed{\dfrac 32}$

@Rishabh Cool , we really liked your comment, and have converted it into a solution. If you subscribe to this solution, you will receive notifications about future comments.

Brilliant Mathematics Staff - 5 years, 3 months ago

Such a brilliant solution!

Harsh Khatri - 5 years, 3 months ago

You forgot to say ... Did the same way (+1).. ;-)

Rishabh Jain - 5 years, 3 months ago

Hahaha! Even I was thinking that I've written it too many times :P

Harsh Khatri - 5 years, 3 months ago

Say No to hospital

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