Question

find
$1) \: \: \displaystyle \lim_{x \to 0} \: \frac{1 - cos (1 - cos \: x )}{ {x}^{4} } \\ \\ 2) \: \: \displaystyle \lim_{x \to 0} \: \frac{ {e}^{ {x}^{2} } - cos \: x }{ {x}^{2} }$
please anyone help me ​

Answer 1

First question :

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{\displaystyle \lim_{x \to 0} \: \frac{1 - cos (1 - cos \: x )}{ {x}^{4} }=\frac{1}{8}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies \displaystyle \lim_{x \to 0} \: \frac{1 - cos (1 - cos \: x )}{ {x}^{4} } \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies \displaystyle \lim_{x \to 0} \: \frac{1 - cos (1 - cos \: x )}{ {x}^{4} } = ?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt : \implies \displaystyle \lim_{x \to 0} \: \frac{1 - cos (1 - cos \: x )}{ {x}^{4} } \\ \\ \tt \circ \: multiply \:and \: divide \: with \: (1 - cos \:x)^{2} \\ \\ \tt: \implies \displaystyle \lim_{x \to 0} \: \frac{1 - cos(1 - cos \: x)}{ {x}^{4} } \times \frac{(1 - cos \: x)^{2} }{(1 - cos \: x)^{2}} \\ \\ \tt : \implies \displaystyle \lim_{x \to 0} \: \frac{1 - cos(1 - cos \: x)}{(1 - cos \: x)^{2}} \times \frac{(1 - cos \: x)^{2}}{ {x}^{4} } \\ \\ \tt \circ \: \displaystyle \lim_{x \to 0} \: \frac{1 - cos \: x}{ {x}^{2} } = \frac{1}{2} \\ \\ \tt: \implies \displaystyle \lim_{x \to 0} \: \frac{1}{2} \times \bigg( \frac{1 - cos \: x}{ {x}^{2} } \bigg)^{2} \\ \\ \tt: \implies \frac{1}{2} \times \bigg(\frac{1}{2} \bigg)^{2} \\ \\ \green{\tt: \implies \frac{1}{8}} \\ \\ \green{\tt\therefore\displaystyle \lim_{x \to 0} \: \frac{1 - cos (1 - cos \: x )}{ {x}^{4} } = \frac{1}{8} }$

Second question :

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{\displaystyle \lim_{x \to 0} \: \frac{e^{x^{2}}- cos \: x}{ {x}^{2} }=\frac{3}{2}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies \displaystyle \lim_{x \to 0} \: \frac{ {e}^{ {x}^{2} } - cos \: x }{ {x}^{2} } \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies \displaystyle \lim_{x \to 0} \: \frac{ {e}^{ {x}^{2} } - cos \: x }{ {x}^{2} } =?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies \displaystyle \lim_{x \to 0} \: \frac{ {e}^{ {x}^{2} } - cos \: x }{ {x}^{2} } \\ \\ \tt: \implies \displaystyle \lim_{x \to 0} \: \frac{ {e}^{ {x}^{2} } - 1 + 1- cos \: x }{ {x}^{2} } \\ \\ \tt: \implies \displaystyle \lim_{x \to 0} \: \bigg(\frac{ {e}^{ {x}^{2} } - 1}{ {x}^{2} } + \frac{1 - cos \: x}{ {x}^{2} } \bigg) \\ \\ \tt \circ \: \displaystyle \lim_{x \to 0} \: \frac{ {e}^{x} - 1 }{ {x}^{2} } = 1 \\ \\ \tt \circ \: \displaystyle \lim_{x \to 0} \: \frac{1 - cos \: x}{ {x}^{2} } = \frac{1}{2} \\ \\ \tt: \implies 1 + \frac{1}{2} \\ \\ \green{\tt: \implies \frac{3}{2}} \\ \\ \green{\tt \therefore \displaystyle \lim_{x \to 0} \: \frac{ {e}^{ {x}^{2} } - cos \: x }{ {x}^{2} } = \frac{3}{2} }$

Answer 2

According to given question :

$</p><p>\begin{gathered}\bold{As \: we \: know \: that} \\ \tt: \implies \displaystyle \lim_{x \to 0} \: \frac{ {e}^{ {x}^{2} } - cos \: x }{ {x}^{2} } \\ \\ \tt: \implies \displaystyle \lim_{x \to 0} \: \frac{ {e}^{ {x}^{2} } - 1 + 1- cos \: x }{ {x}^{2} } \\ \\ \tt: \implies \displaystyle \lim_{x \to 0} \: \bigg(\frac{ {e}^{ {x}^{2} } - 1}{ {x}^{2} } + \frac{1 - cos \: x}{ {x}^{2} } \bigg) \\ \\ \tt \circ \: \displaystyle \lim_{x \to 0} \: \frac{ {e}^{x} - 1 }{ {x}^{2} } = 1 \\ \\ \tt \circ \: \displaystyle \lim_{x \to 0} \: \frac{1 - cos \: x}{ {x}^{2} } = \frac{1}{2} \\ \\ \tt: \implies 1 + \frac{1}{2} \\ \\ \green{\tt: \implies \frac{3}{2}} \\ \\ \green{\tt \therefore \displaystyle \lim_{x \to 0} \: \frac{ {e}^{ {x}^{2} } - cos \: x }{ {x}^{2} } = \frac{3}{2} }\end{gathered} </p><p>Asweknowthat</p><p>:⟹ </p><p>$

$</p><p>x→0</p><p>lim</p><p> </p><p> </p><p>x </p><p>2</p><p> </p><p>e </p><p>x </p><p>2</p><p> </p><p> −cosx</p><p> </p><p> </p><p>:⟹ </p><p>x→0</p><p>lim</p><p> </p><p> </p><p>x </p><p>2</p><p> </p><p>e </p><p>x </p><p>2</p><p> </p><p> −1+1−cosx</p><p> </p><p> </p><p>:⟹ </p><p>x→0</p><p>lim</p><p> </p><p> ( </p><p>x </p><p>2</p><p> </p><p>e </p><p>x </p><p>2</p><p> </p><p> −1</p><p> </p><p> + </p><p>x </p><p>2</p><p> </p><p>1−cosx</p><p> </p><p> )</p><p>∘ </p><p>x→0</p><p>lim</p><p> </p><p> </p><p>x </p><p>2</p><p> </p><p>e </p><p>x</p><p> −1</p><p> </p><p> =1</p><p>∘ </p><p>x→0</p><p>lim</p><p> </p><p> </p><p>x </p><p>2</p><p> </p><p>1−cosx</p><p> </p><p> = </p><p>2</p><p>1</p><p> </p><p> </p><p>:⟹1+ </p><p>2</p><p>1</p><p> </p><p> </p><p>:⟹ </p><p>2</p><p>3</p><p> </p><p> </p><p>∴ </p><p>x→0</p><p>lim</p><p> </p><p> </p><p>x </p><p>2</p><p> </p><p>e </p><p>x </p><p>2</p><p> </p><p> −cosx</p><p> </p><p> = </p><p>2</p><p>3</p><p>$