Question

Solve without using L Hospital Rule

$lim \: x \to \: \frac{\pi}{4} \: \: \frac{sinx - cosx}{x - \frac{\pi}{4} }$
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Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \displaystyle\lim_{x \to \dfrac{\pi}{4} }\rm \frac{sinx - cosx}{x - \dfrac{\pi}{4} }$

Consider,

$\rm \: sinx - cosx$

can be further rewritten as

$\rm \: = \: \sqrt{2}\bigg(\dfrac{1}{ \sqrt{2} }sinx - \dfrac{1}{ \sqrt{2}}cosx \bigg)$

can be further rewritten as

$\rm \: = \: \sqrt{2}\bigg(cos\dfrac{\pi}{4} sinx - sin\dfrac{\pi}{4} cosx \bigg)$

can be re-arranged as

$\rm \: = \: \sqrt{2}\bigg(sinx \: cos\dfrac{\pi}{4} -cosx \: sin\dfrac{\pi}{4} \bigg)$

$\rm \: = \:sin\bigg(x - \dfrac{\pi}{4} \bigg)$

$[\rm \: \because \: sin(x - y) = sinx \: cosy \: - \: siny \: cosx \: \: ]$

So,

$\rm\implies \:sinx - cosx = \sqrt{2}sin\bigg(x - \dfrac{\pi}{4} \bigg)$

So, Substitute this value in the given expression, we get

$\rm \: = \:\displaystyle\lim_{x \to \dfrac{\pi}{4} }\rm \frac{ \sqrt{2} sin\bigg(x - \dfrac{\pi}{4} \bigg) }{x - \dfrac{\pi}{4} }$

can be further rewritten as

$\rm \: = \: \sqrt{2} \: \displaystyle\lim_{x \: - \: \dfrac{\pi}{4} \: \to \: 0}\rm \frac{sin\bigg(x - \dfrac{\pi}{4} \bigg) }{x - \dfrac{\pi}{4} }$

We know,

$\boxed{\sf{  \:\rm \: \displaystyle\lim_{x \to 0}\rm \frac{sinx}{x} = 1 \: \: }}$

So, using this result, we get

$\rm \: = \: \sqrt{2}$

Hence,

$\rm\implies \:\boxed{\sf{  \:\rm \: \displaystyle\lim_{x \to \dfrac{\pi}{4} }\rm \frac{sinx - cosx}{x - \dfrac{\pi}{4} } = \sqrt{2} \: \: }}$

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Additional Information :-

$\boxed{\sf{  \:\rm \: \displaystyle\lim_{x \to 0}\rm \frac{tanx}{x} = 1 \: \: }}$

$\boxed{\sf{  \:\rm \: \displaystyle\lim_{x \to 0}\rm \frac{log(1 + x)}{x} = 1 \: \: }}$

$\boxed{\sf{  \:\rm \: \displaystyle\lim_{x \to 0}\rm \frac{ {e}^{x} - 1}{x} = 1 \: \: }}$

$\boxed{\sf{  \:\rm \: \displaystyle\lim_{x \to 0}\rm \frac{ {a}^{x} - 1}{x} = loga \: \: }}$

Answer 2

Answer:

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