Question

Find Lt x→5 f(x) where f(x) = | x | - 5​
Please Answer Fast

Answer 1

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

Given expression is

$\red{\rm :\longmapsto\:\displaystyle\lim_{x \to 5}f(x) \: \: where \: f(x) = |x| - 5}$

Let first define the function f(x).

We know,

$\begin{gathered}\begin{gathered}\bf\: |x| = \begin{cases} &\sf{ - x \: \: \: \: when \: x < 0} \\ &\sf{ \: \: x \: \: \: \: when \: x \geqslant 0} \end{cases}\end{gathered}\end{gathered}$

So,

$\begin{gathered}\begin{gathered}\bf\: |x| - 5 = \begin{cases} &\sf{ - x - 5 \: \: \: \: when \: x < 0} \\ &\sf{ \: \: x - 5\: \: \: \: when \: x \geqslant 0} \end{cases}\end{gathered}\end{gathered}$

So,

$\begin{gathered}\begin{gathered}\bf\:f(x) = |x| - 5 = \begin{cases} &\sf{ - x - 5 \: \: \: \: when \: x < 0} \\ &\sf{ \: \: x - 5\: \: \: \: when \: x \geqslant 0} \end{cases}\end{gathered}\end{gathered}$

Consider, LHS

$\rm :\longmapsto\:\displaystyle\lim_{x \to 5^{ - }} f(x)$

$\rm \:  =  \: \displaystyle\lim_{x \to 5^{ - }} (x - 5)$

To evaluate this limit, we use method of Substitution.

So, Substitute

$\red{\rm :\longmapsto\:x = 5 - h \: \: as \: x \to \: 5 \: \: so \: h \to \: 0}$

$\rm \:  =  \: \displaystyle\lim_{h \to 0} (5 - h - 5)$

$\rm \:  =  \: \displaystyle\lim_{h \to 0} ( - h )$

$\rm \:  =  \: 0$

So,

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 5^{ - }} f(x) = 0 \: }}$

Now, Consider RHL

$\rm :\longmapsto\:\displaystyle\lim_{x \to 5^{ + }} f(x)$

$\rm \:  =  \: \displaystyle\lim_{x \to 5^{ + }} (x - 5)$

To evaluate this limit, we use method of Substitution.

So, Substitute

$\red{\rm :\longmapsto\:x = 5 + h \: \: as \: x \to \: 5 \: \: so \: h \to \: 0}$

$\rm \:  =  \: \displaystyle\lim_{h \to 0} (5 + h - 5)$

$\rm \:  =  \: \displaystyle\lim_{h \to 0} (h)$

$\rm \:  =  \: 0$

So,

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 5^{ + }} f(x) = 0 \: }}$

So, from above we concluded that

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 5^{ - }} f(x) = \displaystyle\lim_{x \to 5^{ + }} f(x) = 0 \: }}$

So,

$\red{\rm \implies\:\boxed{\tt{ \displaystyle\lim_{x \to 5} f(x) = 0 \: }}}$

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Additional Information :-

$\red{\rm :\longmapsto\:\displaystyle\lim_{x \to 0}\dfrac{sinx}{x} = 1 \: }$

$\red{\rm :\longmapsto\:\displaystyle\lim_{x \to 0}\dfrac{tanx}{x} = 1 \: }$

$\red{\rm :\longmapsto\:\displaystyle\lim_{x \to 0}\dfrac{log(1 + x)}{x} = 1 \: }$

$\red{\rm :\longmapsto\:\displaystyle\lim_{x \to 0}\dfrac{ {e}^{x} - 1}{x} = 1 \: }$

$\red{\rm :\longmapsto\:\displaystyle\lim_{x \to 0}\dfrac{ {a}^{x} - 1}{x} = loga \: }$