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  • Find the value of k so that the function f is continuous at the indicated point: f(x) = root 1 + kx - root 1 - kx

Find the value of k so that the function f is continuous at the indicated point:

\begin{array}{c} f(x )= \frac{\sqrt{1+\mathrm{kx}}-\sqrt{1-\mathrm{kx}}}{\mathrm{x}}, \quad \text { if }-1 \leq \mathrm{x}<0 \\\\ \frac{2 \mathrm{x}+1}{\mathrm{x}-1}, \quad \text { if } 0 \leq \mathrm{x} \leq 1 \end{array}  at x= 0

Answers (1)

Given,

\begin{array}{c} f(x )= \frac{\sqrt{1+\mathrm{kx}}-\sqrt{1-\mathrm{kx}}}{\mathrm{x}}, \quad \text { if }-1 \leq \mathrm{x}<0 \\\\ \frac{2 \mathrm{x}+1}{\mathrm{x}-1}, \quad \text { if } 0 \leq \mathrm{x} \leq 1 \end{array}

We need to find the value of k such that f(x) is continuous at x = 0.
A function f(x) is said to be continuous at x = c if,
Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).
Mathematically we can represent it as-

\lim _{h \rightarrow 0} f(c-h)=\lim _{h \rightarrow 0} f(c+h)=f(c)

Where h is a very small number very close to 0 (h→0)

Now, let’s assume that f(x) is continuous at x = 0.

\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(0+h)=f(0)

Now to find k pick out a combination using which we get k in our equation.
In this question we take LHL = f(0)

\begin{array}{l} \quad \lim _{h \rightarrow 0} f(-h)=f(0) \\ \Rightarrow \lim _{h \rightarrow 0}\left\{\frac{\sqrt{1+k(-h)}-\sqrt{1-k(-h)})}{-h}\right\}=\frac{2(0)+1}{(0)-1}\{\text { using eqn } 1\} \\ \Rightarrow \lim _{h \rightarrow 0}\left\{\frac{\sqrt{1+k h}-\sqrt{1-k h}}{h}\right\}=-1 \end{array}

We can’t find the limit directly, because it is taking 0/0 form.
thus, we will rationalize it.

\begin{aligned} &\Rightarrow \lim _{h \rightarrow 0}\left\{\frac{\sqrt{1+k h}-\sqrt{1-k h}}{h} \times \frac{\sqrt{1+k h}+\sqrt{1-k h}}{\sqrt{1+k h}+\sqrt{1-k h}}\right\}=-1\\ &\text { Using }(a+b)(a-b)=a^{2}-b^{2}, \text { we have }-\\ &\lim _{h \rightarrow 0}\left\{\frac{(\sqrt{1+k h})^{2}-(\sqrt{1-k h)})^{2}}{h(\sqrt{1+k h}+\sqrt{1-k h})}\right\}=-1\\ &\Rightarrow \lim _{h \rightarrow 0}\left\{\frac{2 k h}{h(\sqrt{1+k h}+\sqrt{1-k h})}\right\}=-1\\ &\lim _{h \rightarrow 0}\left\{\frac{2 k}{(\sqrt{1+k h}+\sqrt{1-k h})}\right\}=\\ &\Rightarrow \frac{2 k}{\sqrt{1+k(0)}+\sqrt{1-k(0)}}=-1\\ &\therefore 2 k / 2=-1 \\&\therefore k =-1 \end{aligned}

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