Limit the Limit....

Calculus Level 3

Evaluate

$\lim _{ x\rightarrow 0 }{ \frac { { e }^{ { x }^{ 2 } }-\cos { (x) } }{ { x }^{ 2 } } }$

For more challenge: Try doing it without using L'Hospital's Rule

The answer is 1.5.

1 solution

Ankush Tiwari
Jun 22, 2014

$\lim_{x \to 0} \frac{{ e }^{ { x }^{ 2 } } - cosx}{x^2}$

= $\lim_{x \to 0} \frac{({ e }^{ { x }^{ 2 } }-1) +(1- cosx)}{x^2}$

= $\lim_{x \to 0} \frac{({ e }^{ { x }^{ 2 } }-1)}{x^2}+\lim_{x \to 0} \frac{(1- cosx)}{x^2}$ = $1 + \frac{1}{2}$ = $1.5$

Can you explain to us how you did it without L'Hopital's Rule?

Kenny Lau - 6 years, 11 months ago

He simply added a special 0 value and used identities....

Matt Schramm - 6 years, 10 months ago

The second limit which you evaluated as $\frac{1}{2}$ , did you use any special identity that I don't know of? I did it in the same way as you but I used the identities $(1-\cos x)=2\sin^2 (\frac{x}{2})$ and $\displaystyle \lim_{y\to 0} \frac{\sin y}{y}=1$ to evaluate the second one. The first part, I did the same as you.

Prasun Biswas - 6 years, 6 months ago

what did you considered " y" to be , x^{2} / 2 ryt?

Anurag Pandey - 6 years, 1 month ago

Nope, I considered $y=\dfrac{x}{2}$ .

Prasun Biswas - 6 years, 1 month ago

@Prasun Biswas – okay. Then there must be a square outside it ryt?

Anurag Pandey - 6 years, 1 month ago

@Anurag Pandey – Yes, you're right in that interpretation. :)

Prasun Biswas - 6 years, 1 month ago

Limit the Limit....

The answer is 1.5.

1 solution

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